Ygthkb

Limiting value of a sequence when n tends to infinity

Minimum Viable Product for RTS game?

How to deal with an underperforming subordinate?

Are all power cords made equal?

Including proofs of known theorems in master's thesis

Why is it that Bernie Sanders is always called a "socialist"?

How long can the stop in a stop-and-go be?

Why is Shelob considered evil?

I am a giant among ants

Protagonist constantly has to have long words explained to her. Will this get tedious?

Third wheel character

If a 12 by 16 sheet of paper is folded on its diagonal, what is the area of the region of the overlap?

What do "compile" , "fit" and "predict" do in Keras sequential models?

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

What's the reason that we have different quantities of days each month?

How do I fight with Heavy Armor as a Wizard with Tenser's Transformation?

Why did Ylvis use "go" instead of "say" in phrases like "Dog goes 'woof'"?

Isn't a semicolon (';') needed after a function declaration in C++?

Why does a single AND gate need 60 transistors?

Coworker asking me to not bring cakes due to self control issue. What should I do?

How to write Muḥammad ibn Mūsā al-Khwārizmī?

Is practicing on a digital piano harmful to an experienced piano player?

Why do single electrical receptacles exist?

Taking out the plank from one's own eye

Limiting value of a sequence when n tends to infinity

How do I evaluate $prod_{r=1}^{infty }left (1-frac{1}{sqrt {r+1}}right)$?Find the value of $lim_{n rightarrow infty} Big( 1-frac{1}{sqrt 2} Big) cdots Big(1-frac{1}{sqrt {n+1}} Big)$Let $a_{n}=(1-frac{1}{sqrt{2}})ldots(1-frac{1}{sqrt{n+1}}), n geq 1 $. Then $lim_limits{n rightarrow infty} a_{n} $If $a_n=left(1-frac{1}{sqrt{2}}right)ldotsleft(1-frac{1}{sqrt{n+1}}right)$ then $lim_{ntoinfty}a_n=?$How can I find $lim_{nto infty} a_n$Finding the limit of the sequence $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt {n+1})$$mathbf{a_n = (1-frac{1}{sqrt2})…(1-frac{1}{sqrt{n+1}})}$, $n ge 1$ $Rightarrow$ $lim_{nto infty} a_n$ =?Intersection of 3 planes in 3D spaceWhy prime number theorem tends to one?Constant sequence limit at infinityMaximum value of a sequencehow to simplify $frac{d}{dx}left(ln left(sqrt{frac{1}{2-x}}right)right)$Find the Limit of the given sequence x_n = $(1 - 1/3 )^2$ $(1 - 1/6)^2$ $(1 - 1/10)^2$…$(1 - 2/n(n+1))^2$, n>=2find $lim_{ntoinfty}(1+frac{1}{3})(1+frac{1}{3^2})(1+frac{1}{3^4})cdots(1+frac{1}{3^{2^n}})$limits to infinity when x goes to infinityLimiting value of this expressionFind the limiting value of $S=a^{sqrt{1}}+a^{sqrt{2}}+a^{sqrt{3}}+a^{sqrt{4}}+…$ for $0 leq a < 1$Find the limit of this as n tends to infinity

1

$begingroup$

Q) Let, $a_{n} ;=; left ( 1-frac{1}{sqrt{2}} right ) ... left ( 1- frac{1}{sqrt{n+1}} right )$ , $n geq 1$. Then $lim_{nrightarrow infty } a_{n}$

(A) equals $1$

(B) does not exist

(C) equals $frac{1}{sqrt{pi }}$

(D) equals $0$

My Approach :- I am not getting a particular direction or any procedure to simplify $a_{n}$ and find its value when n tends to infinity.

So, I tried like this simple way to substitute values and trying to find the limiting value :-
$left ( 1-frac{1}{sqrt{1+1}} right ) * left ( 1-frac{1}{sqrt{2+1}} right )*left ( 1-frac{1}{sqrt{3+1}} right )*left ( 1-frac{1}{sqrt{4+1}} right )*left ( 1-frac{1}{sqrt{5+1}} right )*left ( 1-frac{1}{sqrt{6+1}} right )*left ( 1-frac{1}{sqrt{7+1}} right )*left ( 1-frac{1}{sqrt{8+1}} right )*.........*left ( 1-frac{1}{sqrt{n+1}} right )$

=$(0.293)*(0.423)*(0.5)*(0.553)*(0.622)*(0.647)*(0.667)* ....$
=0.009*...

So, here value is tending to zero. I think option $(D)$ is correct.

I have tried like this
$left ( frac{sqrt{2}-1}{sqrt{2}} right )*left ( frac{sqrt{3}-1}{sqrt{3}} right )*left ( frac{sqrt{4}-1}{sqrt{4}} right )*.......left ( frac{sqrt{(n+1)}-1}{sqrt{n+1}} right )$

= $left ( frac{(sqrt{2}-1)*(sqrt{3}-1)*(sqrt{4}-1)*.......*(sqrt{n+1}-1)}{{sqrt{(n+1)!}}} right )$

Now, again I stuck how to simplify further and find the value for which $a_{n}$ converges when $n$ tends to infinity . Please help if there is any procedure to solve this question.

calculus sequences-and-series limits products

share|cite|improve this question

edited 12 mins ago

Michael Rozenberg

105k1892197

asked 35 mins ago

ankitankit

175

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Using the inequality $1+x leq e^x$ which is true for all $x in mathbb{R}$, we have $$ a_n leq expleft{ -sum_{k=2}^{n} frac{1}{sqrt{k}} right} leq C expleft{ -int_{0}^{n} frac{mathrm{d}x}{sqrt{x}} right} = C e^{-2sqrt{n}} $$ for some constant $C > 0$. From this, we can conclude $a_n to 0$. A more natural approach is to take log and utilize Taylor approximation, which in turn yields $a_n = e^{-2sqrt{n} + mathcal{O}(1)}$ as $ntoinfty$.
  $endgroup$
  – Sangchul Lee
  30 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What would you do if it were $$left(1-frac12right)left(1-frac13right)cdotsleft(1-frac1{n+1}right)?$$
  $endgroup$
  – Lord Shark the Unknown
  29 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Been asked here couple of times: $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt {n+1})$, $mathbf{a_n = (1-frac{1}{sqrt2})...(1-frac{1}{sqrt{n+1}})}$, $n ge 1$ $Rightarrow$ $lim_{nto infty} a_n$ =?, Find the value of $lim_{n rightarrow infty} Big( 1-frac{1}{sqrt 2} Big) cdots Big(1-frac{1}{sqrt {n+1}} Big)$,
  $endgroup$
  – Sil
  10 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $a_n=left(1-frac{1}{sqrt{2}}right)ldotsleft(1-frac{1}{sqrt{n+1}}right)$ then $lim_{ntoinfty}a_n=?$, How can I find $lim_{nto infty} a_n$ and Let $a_{n}=(1-frac{1}{sqrt{2}})ldots(1-frac{1}{sqrt{n+1}}), n geq 1 $. Then $lim_limits{n rightarrow infty} a_{n} $...
  $endgroup$
  – Sil
  10 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of How can I find $lim_{nto infty} a_n$
  $endgroup$
  – Sil
  8 mins ago
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

Q) Let, $a_{n} ;=; left ( 1-frac{1}{sqrt{2}} right ) ... left ( 1- frac{1}{sqrt{n+1}} right )$ , $n geq 1$. Then $lim_{nrightarrow infty } a_{n}$

(A) equals $1$

(B) does not exist

(C) equals $frac{1}{sqrt{pi }}$

(D) equals $0$

My Approach :- I am not getting a particular direction or any procedure to simplify $a_{n}$ and find its value when n tends to infinity.

So, I tried like this simple way to substitute values and trying to find the limiting value :-
$left ( 1-frac{1}{sqrt{1+1}} right ) * left ( 1-frac{1}{sqrt{2+1}} right )*left ( 1-frac{1}{sqrt{3+1}} right )*left ( 1-frac{1}{sqrt{4+1}} right )*left ( 1-frac{1}{sqrt{5+1}} right )*left ( 1-frac{1}{sqrt{6+1}} right )*left ( 1-frac{1}{sqrt{7+1}} right )*left ( 1-frac{1}{sqrt{8+1}} right )*.........*left ( 1-frac{1}{sqrt{n+1}} right )$

=$(0.293)*(0.423)*(0.5)*(0.553)*(0.622)*(0.647)*(0.667)* ....$
=0.009*...

So, here value is tending to zero. I think option $(D)$ is correct.

I have tried like this
$left ( frac{sqrt{2}-1}{sqrt{2}} right )*left ( frac{sqrt{3}-1}{sqrt{3}} right )*left ( frac{sqrt{4}-1}{sqrt{4}} right )*.......left ( frac{sqrt{(n+1)}-1}{sqrt{n+1}} right )$

= $left ( frac{(sqrt{2}-1)*(sqrt{3}-1)*(sqrt{4}-1)*.......*(sqrt{n+1}-1)}{{sqrt{(n+1)!}}} right )$

Now, again I stuck how to simplify further and find the value for which $a_{n}$ converges when $n$ tends to infinity . Please help if there is any procedure to solve this question.

calculus sequences-and-series limits products

share|cite|improve this question

edited 12 mins ago

Michael Rozenberg

105k1892197

asked 35 mins ago

ankitankit

175

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Using the inequality $1+x leq e^x$ which is true for all $x in mathbb{R}$, we have $$ a_n leq expleft{ -sum_{k=2}^{n} frac{1}{sqrt{k}} right} leq C expleft{ -int_{0}^{n} frac{mathrm{d}x}{sqrt{x}} right} = C e^{-2sqrt{n}} $$ for some constant $C > 0$. From this, we can conclude $a_n to 0$. A more natural approach is to take log and utilize Taylor approximation, which in turn yields $a_n = e^{-2sqrt{n} + mathcal{O}(1)}$ as $ntoinfty$.
  $endgroup$
  – Sangchul Lee
  30 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What would you do if it were $$left(1-frac12right)left(1-frac13right)cdotsleft(1-frac1{n+1}right)?$$
  $endgroup$
  – Lord Shark the Unknown
  29 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Been asked here couple of times: $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt {n+1})$, $mathbf{a_n = (1-frac{1}{sqrt2})...(1-frac{1}{sqrt{n+1}})}$, $n ge 1$ $Rightarrow$ $lim_{nto infty} a_n$ =?, Find the value of $lim_{n rightarrow infty} Big( 1-frac{1}{sqrt 2} Big) cdots Big(1-frac{1}{sqrt {n+1}} Big)$,
  $endgroup$
  – Sil
  10 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $a_n=left(1-frac{1}{sqrt{2}}right)ldotsleft(1-frac{1}{sqrt{n+1}}right)$ then $lim_{ntoinfty}a_n=?$, How can I find $lim_{nto infty} a_n$ and Let $a_{n}=(1-frac{1}{sqrt{2}})ldots(1-frac{1}{sqrt{n+1}}), n geq 1 $. Then $lim_limits{n rightarrow infty} a_{n} $...
  $endgroup$
  – Sil
  10 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of How can I find $lim_{nto infty} a_n$
  $endgroup$
  – Sil
  8 mins ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Q) Let, $a_{n} ;=; left ( 1-frac{1}{sqrt{2}} right ) ... left ( 1- frac{1}{sqrt{n+1}} right )$ , $n geq 1$. Then $lim_{nrightarrow infty } a_{n}$

(A) equals $1$

(B) does not exist

(C) equals $frac{1}{sqrt{pi }}$

(D) equals $0$

My Approach :- I am not getting a particular direction or any procedure to simplify $a_{n}$ and find its value when n tends to infinity.

So, I tried like this simple way to substitute values and trying to find the limiting value :-
$left ( 1-frac{1}{sqrt{1+1}} right ) * left ( 1-frac{1}{sqrt{2+1}} right )*left ( 1-frac{1}{sqrt{3+1}} right )*left ( 1-frac{1}{sqrt{4+1}} right )*left ( 1-frac{1}{sqrt{5+1}} right )*left ( 1-frac{1}{sqrt{6+1}} right )*left ( 1-frac{1}{sqrt{7+1}} right )*left ( 1-frac{1}{sqrt{8+1}} right )*.........*left ( 1-frac{1}{sqrt{n+1}} right )$

=$(0.293)*(0.423)*(0.5)*(0.553)*(0.622)*(0.647)*(0.667)* ....$
=0.009*...

So, here value is tending to zero. I think option $(D)$ is correct.

I have tried like this
$left ( frac{sqrt{2}-1}{sqrt{2}} right )*left ( frac{sqrt{3}-1}{sqrt{3}} right )*left ( frac{sqrt{4}-1}{sqrt{4}} right )*.......left ( frac{sqrt{(n+1)}-1}{sqrt{n+1}} right )$

= $left ( frac{(sqrt{2}-1)*(sqrt{3}-1)*(sqrt{4}-1)*.......*(sqrt{n+1}-1)}{{sqrt{(n+1)!}}} right )$

Now, again I stuck how to simplify further and find the value for which $a_{n}$ converges when $n$ tends to infinity . Please help if there is any procedure to solve this question.

calculus sequences-and-series limits products

share|cite|improve this question

edited 12 mins ago

Michael Rozenberg

105k1892197

asked 35 mins ago

ankitankit

175

$endgroup$

share|cite|improve this question

edited 12 mins ago

Michael Rozenberg

105k1892197

asked 35 mins ago

ankitankit

175

share|cite|improve this question

edited 12 mins ago

Michael Rozenberg

105k1892197

asked 35 mins ago

ankitankit

175

edited 12 mins ago

Michael Rozenberg

105k1892197

edited 12 mins ago

Michael Rozenberg

105k1892197

asked 35 mins ago

ankitankit

175

asked 35 mins ago

ankitankit

175

2 Answers
2

active

oldest

votes

3

$begingroup$

The hint:
$$0<a_n=prod_{k=1}^nleft(1-frac{1}{sqrt{k+1}}right)<prod_{k=1}^nleft(1-frac{1}{k+1}right)=frac{1}{n+1}rightarrow0$$

share|cite|improve this answer

answered 25 mins ago

Michael RozenbergMichael Rozenberg

105k1892197

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Michael.In 2 lines! :)
  $endgroup$
  – Peter Szilas
  14 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

As we multiply positive factors below $1$, the sequence is positive and strictly decreasing, hence convergent (ruling out B). As already $a_1<1$, we also rule out A.

If we multiply $a_n$ by $b_n:=left(1+frac1{sqrt 2}right)cdots left(1+frac1{sqrt {n+1}}right)$, note that the product of corresponding factors is $left(1-frac1{sqrt {k}}right)left(1+frac1{sqrt {k}}right)=1-frac1k<1$, hence $a_nb_n<1$. On the other hand, by expanding the product and dropping lots of positive terms
$$b_nge 1+frac1{sqrt 2}+frac1{sqrt 3}+ldots+frac1{sqrt {n+1}} $$
so that $b_n$ gets arbitrarily large. We conclude that $a_n$ gets arbitrarily small positive. In other words $a_nto 0$.

share|cite|improve this answer

answered 25 mins ago

Hagen von EitzenHagen von Eitzen

280k23272505

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hagen.Very nice.
  $endgroup$
  – Peter Szilas
  15 mins ago
  

  
  
  
  

add a comment | 

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

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3124615%2flimiting-value-of-a-sequence-when-n-tends-to-infinity%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

The hint:
$$0<a_n=prod_{k=1}^nleft(1-frac{1}{sqrt{k+1}}right)<prod_{k=1}^nleft(1-frac{1}{k+1}right)=frac{1}{n+1}rightarrow0$$

share|cite|improve this answer

answered 25 mins ago

Michael RozenbergMichael Rozenberg

105k1892197

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Michael.In 2 lines! :)
  $endgroup$
  – Peter Szilas
  14 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

The hint:
$$0<a_n=prod_{k=1}^nleft(1-frac{1}{sqrt{k+1}}right)<prod_{k=1}^nleft(1-frac{1}{k+1}right)=frac{1}{n+1}rightarrow0$$

share|cite|improve this answer

answered 25 mins ago

Michael RozenbergMichael Rozenberg

105k1892197

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Michael.In 2 lines! :)
  $endgroup$
  – Peter Szilas
  14 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

The hint:
$$0<a_n=prod_{k=1}^nleft(1-frac{1}{sqrt{k+1}}right)<prod_{k=1}^nleft(1-frac{1}{k+1}right)=frac{1}{n+1}rightarrow0$$

share|cite|improve this answer

answered 25 mins ago

Michael RozenbergMichael Rozenberg

105k1892197

$endgroup$

share|cite|improve this answer

answered 25 mins ago

Michael RozenbergMichael Rozenberg

105k1892197

answered 25 mins ago

Michael RozenbergMichael Rozenberg

105k1892197

answered 25 mins ago

Michael RozenbergMichael Rozenberg

105k1892197

2

$begingroup$

As we multiply positive factors below $1$, the sequence is positive and strictly decreasing, hence convergent (ruling out B). As already $a_1<1$, we also rule out A.

If we multiply $a_n$ by $b_n:=left(1+frac1{sqrt 2}right)cdots left(1+frac1{sqrt {n+1}}right)$, note that the product of corresponding factors is $left(1-frac1{sqrt {k}}right)left(1+frac1{sqrt {k}}right)=1-frac1k<1$, hence $a_nb_n<1$. On the other hand, by expanding the product and dropping lots of positive terms
$$b_nge 1+frac1{sqrt 2}+frac1{sqrt 3}+ldots+frac1{sqrt {n+1}} $$
so that $b_n$ gets arbitrarily large. We conclude that $a_n$ gets arbitrarily small positive. In other words $a_nto 0$.

share|cite|improve this answer

answered 25 mins ago

Hagen von EitzenHagen von Eitzen

280k23272505

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hagen.Very nice.
  $endgroup$
  – Peter Szilas
  15 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

As we multiply positive factors below $1$, the sequence is positive and strictly decreasing, hence convergent (ruling out B). As already $a_1<1$, we also rule out A.

If we multiply $a_n$ by $b_n:=left(1+frac1{sqrt 2}right)cdots left(1+frac1{sqrt {n+1}}right)$, note that the product of corresponding factors is $left(1-frac1{sqrt {k}}right)left(1+frac1{sqrt {k}}right)=1-frac1k<1$, hence $a_nb_n<1$. On the other hand, by expanding the product and dropping lots of positive terms
$$b_nge 1+frac1{sqrt 2}+frac1{sqrt 3}+ldots+frac1{sqrt {n+1}} $$
so that $b_n$ gets arbitrarily large. We conclude that $a_n$ gets arbitrarily small positive. In other words $a_nto 0$.

share|cite|improve this answer

answered 25 mins ago

Hagen von EitzenHagen von Eitzen

280k23272505

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hagen.Very nice.
  $endgroup$
  – Peter Szilas
  15 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

As we multiply positive factors below $1$, the sequence is positive and strictly decreasing, hence convergent (ruling out B). As already $a_1<1$, we also rule out A.

If we multiply $a_n$ by $b_n:=left(1+frac1{sqrt 2}right)cdots left(1+frac1{sqrt {n+1}}right)$, note that the product of corresponding factors is $left(1-frac1{sqrt {k}}right)left(1+frac1{sqrt {k}}right)=1-frac1k<1$, hence $a_nb_n<1$. On the other hand, by expanding the product and dropping lots of positive terms
$$b_nge 1+frac1{sqrt 2}+frac1{sqrt 3}+ldots+frac1{sqrt {n+1}} $$
so that $b_n$ gets arbitrarily large. We conclude that $a_n$ gets arbitrarily small positive. In other words $a_nto 0$.

share|cite|improve this answer

answered 25 mins ago

Hagen von EitzenHagen von Eitzen

280k23272505

$endgroup$

share|cite|improve this answer

answered 25 mins ago

Hagen von EitzenHagen von Eitzen

280k23272505

answered 25 mins ago

Hagen von EitzenHagen von Eitzen

280k23272505

answered 25 mins ago

Hagen von EitzenHagen von Eitzen

280k23272505