How do you determine if the following series converges?Convergence and exponentialsDetermine if the series...

Does しかたない imply disappointment?

What is this mysterious *green square* on my Content Editor?

Is Developer Console going to be deprecated?

Partial derivative with respect to three variables

How does holding onto an active but un-used credit card affect your ability to get a loan?

How can I deal with my coworker having zero social cues?

Buying a "Used" Router

How do I make my single-minded character more interested in the main story?

What really causes series inductance of capacitors?

How to transport 10,000 terrestrial trolls across ocean fast?

How bad is a Computer Science course that doesn't teach Design Patterns?

What does "don't have a baby" imply or mean in this sentence?

Minimum Viable Product for RTS game?

Including proofs of known theorems in master's thesis

How to deal with an underperforming subordinate?

Can an attached stirge deal no more than 10 HP of damage before it detaches?

If I tried and failed to start my own business, how do I apply for a job without job experience?

Is there any way to play D&D without a DM?

Was the Spartan by Mimic Systems a real product?

Missing a connection and don't have money to book next flight

Taking an academic pseudonym?

Expression for "unconsciously using words (or accents) used by a person you often talk with or listen to"?

Graphical modeler calculator missing in QGIS3

How can I handle players killing my NPC outside of combat?



How do you determine if the following series converges?


Convergence and exponentialsDetermine if the series converges/divergesHow to prove this series converges $sum_{n=2}^infty frac{ln(n)}{n^{3/2}}$?How to determine if this series converges?Divergence test for the series $sum_{n=1}^{infty} (sqrt[n]{2n^2}-1)^n$Determine whether the series converges or diverges?My approach to determine if the following series is convergentHow to prove that $sum_{n=1}^{infty} frac{(log (n))^2}{n^2}$ converges?Determine whether the series converges or diverges.Does $displaystylesum_{n=1}^{infty}sinleft(displaystylefrac{1}{sqrt{n}}right)$ converge?Find all values of $k$ such that the series with terms $k^n / n^k$ converges.













3












$begingroup$


$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?






convergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    37 mins ago










  • $begingroup$
    Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
    $endgroup$
    – JavaMan
    31 mins ago












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    1 min ago
















3












$begingroup$


$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?






convergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    37 mins ago










  • $begingroup$
    Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
    $endgroup$
    – JavaMan
    31 mins ago












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    1 min ago














3












3








3





$begingroup$


$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?






convergence







share|cite|improve this question









$endgroup$




$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?






convergence




convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 48 mins ago









JayJay

334




334












  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    37 mins ago










  • $begingroup$
    Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
    $endgroup$
    – JavaMan
    31 mins ago












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    1 min ago


















  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    37 mins ago










  • $begingroup$
    Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
    $endgroup$
    – JavaMan
    31 mins ago












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    1 min ago
















$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
37 mins ago




$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
37 mins ago












$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
31 mins ago






$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
31 mins ago














$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
1 min ago




$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
1 min ago










4 Answers
4






active

oldest

votes


















0












$begingroup$

HINT:



Note that $$left( 1-frac1k right)^kle e^{-1}$$



Can you finish?






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    The root test works. Consider
    $$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
    hence the series converges.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      The ratio test is also interesting
      $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
      $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



      Develop as a Taylor series for large values of $k$ to get
      $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
      Continue with Taylor
      $$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        $begin{array}\
        (1-frac1{k})^{k^2}
        &=(frac{k-1}{k})^{k^2}\
        &=dfrac1{(frac{k}{k-1})^{k^2}}\
        &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
        &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
        &<dfrac1{(1+frac{k}{k-1})^k}
        qquadtext{by Bernoulli}\
        &=dfrac1{(frac{2k-1}{k-1})^k}\
        &<dfrac1{(frac{2k-2}{k-1})^k}\
        &=dfrac1{2^k}\
        end{array}
        $



        and the sum of this converges.






        share|cite|improve this answer









        $endgroup$













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3123320%2fhow-do-you-determine-if-the-following-series-converges%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          HINT:



          Note that $$left( 1-frac1k right)^kle e^{-1}$$



          Can you finish?






          share|cite|improve this answer









          $endgroup$


















            0












            $begingroup$

            HINT:



            Note that $$left( 1-frac1k right)^kle e^{-1}$$



            Can you finish?






            share|cite|improve this answer









            $endgroup$
















              0












              0








              0





              $begingroup$

              HINT:



              Note that $$left( 1-frac1k right)^kle e^{-1}$$



              Can you finish?






              share|cite|improve this answer









              $endgroup$



              HINT:



              Note that $$left( 1-frac1k right)^kle e^{-1}$$



              Can you finish?







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 33 mins ago









              Mark ViolaMark Viola

              132k1277174




              132k1277174























                  3












                  $begingroup$

                  The root test works. Consider
                  $$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
                  hence the series converges.






                  share|cite|improve this answer









                  $endgroup$


















                    3












                    $begingroup$

                    The root test works. Consider
                    $$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
                    hence the series converges.






                    share|cite|improve this answer









                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      The root test works. Consider
                      $$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
                      hence the series converges.






                      share|cite|improve this answer









                      $endgroup$



                      The root test works. Consider
                      $$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
                      hence the series converges.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 37 mins ago









                      Theo BenditTheo Bendit

                      18.7k12253




                      18.7k12253























                          0












                          $begingroup$

                          The ratio test is also interesting
                          $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
                          $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



                          Develop as a Taylor series for large values of $k$ to get
                          $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
                          Continue with Taylor
                          $$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            The ratio test is also interesting
                            $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
                            $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



                            Develop as a Taylor series for large values of $k$ to get
                            $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
                            Continue with Taylor
                            $$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              The ratio test is also interesting
                              $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
                              $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



                              Develop as a Taylor series for large values of $k$ to get
                              $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
                              Continue with Taylor
                              $$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$






                              share|cite|improve this answer









                              $endgroup$



                              The ratio test is also interesting
                              $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
                              $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



                              Develop as a Taylor series for large values of $k$ to get
                              $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
                              Continue with Taylor
                              $$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 24 mins ago









                              Claude LeiboviciClaude Leibovici

                              122k1157134




                              122k1157134























                                  0












                                  $begingroup$

                                  $begin{array}\
                                  (1-frac1{k})^{k^2}
                                  &=(frac{k-1}{k})^{k^2}\
                                  &=dfrac1{(frac{k}{k-1})^{k^2}}\
                                  &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
                                  &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
                                  &<dfrac1{(1+frac{k}{k-1})^k}
                                  qquadtext{by Bernoulli}\
                                  &=dfrac1{(frac{2k-1}{k-1})^k}\
                                  &<dfrac1{(frac{2k-2}{k-1})^k}\
                                  &=dfrac1{2^k}\
                                  end{array}
                                  $



                                  and the sum of this converges.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    $begin{array}\
                                    (1-frac1{k})^{k^2}
                                    &=(frac{k-1}{k})^{k^2}\
                                    &=dfrac1{(frac{k}{k-1})^{k^2}}\
                                    &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
                                    &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
                                    &<dfrac1{(1+frac{k}{k-1})^k}
                                    qquadtext{by Bernoulli}\
                                    &=dfrac1{(frac{2k-1}{k-1})^k}\
                                    &<dfrac1{(frac{2k-2}{k-1})^k}\
                                    &=dfrac1{2^k}\
                                    end{array}
                                    $



                                    and the sum of this converges.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      $begin{array}\
                                      (1-frac1{k})^{k^2}
                                      &=(frac{k-1}{k})^{k^2}\
                                      &=dfrac1{(frac{k}{k-1})^{k^2}}\
                                      &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
                                      &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
                                      &<dfrac1{(1+frac{k}{k-1})^k}
                                      qquadtext{by Bernoulli}\
                                      &=dfrac1{(frac{2k-1}{k-1})^k}\
                                      &<dfrac1{(frac{2k-2}{k-1})^k}\
                                      &=dfrac1{2^k}\
                                      end{array}
                                      $



                                      and the sum of this converges.






                                      share|cite|improve this answer









                                      $endgroup$



                                      $begin{array}\
                                      (1-frac1{k})^{k^2}
                                      &=(frac{k-1}{k})^{k^2}\
                                      &=dfrac1{(frac{k}{k-1})^{k^2}}\
                                      &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
                                      &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
                                      &<dfrac1{(1+frac{k}{k-1})^k}
                                      qquadtext{by Bernoulli}\
                                      &=dfrac1{(frac{2k-1}{k-1})^k}\
                                      &<dfrac1{(frac{2k-2}{k-1})^k}\
                                      &=dfrac1{2^k}\
                                      end{array}
                                      $



                                      and the sum of this converges.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 13 mins ago









                                      marty cohenmarty cohen

                                      73.8k549128




                                      73.8k549128






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3123320%2fhow-do-you-determine-if-the-following-series-converges%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          -

                                          Popular posts from this blog

                                          Discografia di Klaus Schulze Indice Album in studio | Album dal vivo | Singoli | Antologie | Colonne...

                                          Image
                                          Discografie di artisti tedeschi Voce principale: Klaus Schulze . Discografia di Klaus Schulze Klaus Schulze e Lisa Gerrard durante un concerto Album in studio 56 ↙ Album dal vivo 15 ↙ Raccolte 37 ↙ Singoli 4 ↙ Album video 1 ↙ Colonne sonore 4 ↙ Indice 1 Album in studio 1.1 Solisti 1.1.1 Come Richard Wahnfried 1.2 Collaborazioni 1.2.1 Con Rainer Bloss 1.2.2 Con Rainer Bloss ed Ernst Fuchs 1.2.3 Con Lisa Gerrard 1.2.4 Con Andreas Grosser 1.2.5 Con Pete Namlook 1.2.6 Con Pete Namlook e Bill Laswell 1.2.7 Con Michael Shrieve 1.2.8 Con Solar Moon 1.2.9 Con artisti vari 2 Album dal vivo 3 Singoli 4 Antologie 5 Colonne sonore 6 Videografia 7 Collegamenti musica Album in studio | Solisti | 1972 – Irrlicht 1973 – Cyborg 1974 – Blackdance 1975 – Timewind 1975 – Picture Music 1976 – Moondawn 1977 – Mirage 1978 – "X" 1979 – Dune...
                                          Read more

                                          Armoriale delle famiglie italiane (Car) Indice Armi | Bibliografia | Menu di navigazioneBlasone...

                                          Image
                                          Armoriali delle famiglie italiane Blasone Cesenate Famiglie nobili di SiciliaFamiglie nobili di SiciliaTropea MagazineTropea MagazineNobili napoletani Nobili napoletani[1]Nobili napoletaniNobili napoletaniTropea Magazine Nobili napoletaniNobili napoletani Famiglie nobili di SiciliaBlasonario CremoneseFamiglie nobili di Sicilia Archivio Storico CapitolinoRaccolta stemmi TrippiniFamiglie nobili di SiciliaNobili napoletaniArmas de los cavalleros de Veneçia Tropea Magazine Famiglie nobili di Sicilia Nobili napoletaniFamiglie nobili di SiciliaRaccolta stemmi Topoguz Famiglie nobili di SiciliaBlasone CesenateRaccolta stemmi TrippiniArchivio Storico Capitolino Famiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di Sicilia Famiglie nobili di SiciliaFamiglie nobili di SiciliaBlasone Cesenate Raccolta stemmi TrippiniFamiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di Sicilia Blasonario CremoneseBlasonario Cr...
                                          Read more

                                          Lupi Siderali Indice Storia | Organizzazione | La Tredicesima Compagnia | Aspetto | Membri Importanti...

                                          Capitoli degli Space Marine Space MarineImperium della Terrawargame tridimensionaleWarhammer 40.000PrimarcalupiuroImperatoreRune di DissimulazioneAngeli OscuriEresia di HorusStirpe dei MilleOcchio del TerroreCaosTzeentchHorusCodex AstartesSeconda FondazioneFratelli del LupoOcchio del TerroreAlbero della VitareliquieGuardia ImperialeLogan Grimnartruppe d'assaltoImperium dell'UomoveteraniStirpe dei MilleCaosOcchio del TerroreAngeli SanguinariEresia di Horus Lupi Siderali Nome originale Space Wolves Primarca Leman Russ Grido di battaglia Per Russ e per il Padre di Tutti (un feroce ululato) Colori Grigio Altri capitoli I Lupi Siderali sono una delle venti Legioni della Prima Fondazione degli Space Marine che servono l'Imperium della Terra nel wargame tridimensionale Warhammer 40.000. Sono formidabili nel corpo a corpo, i Confratelli desiderano solo la gloria sul campo di battaglia. Indice 1 Storia 1.1 Leman Russ 1.2 La ...
                                          Read more