Limiting value of a sequence when n tends to infinity [duplicate]
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This question already has an answer here:
How can I find $lim_nto infty a_n$ [duplicate]
2 answers
How do I evaluate $prod_r=1^infty left (1-frac1sqrt r+1right)$?
2 answers
If $a_n=left(1-frac1sqrt2right)ldotsleft(1-frac1sqrtn+1right)$ then $lim_ntoinftya_n=?$ [duplicate]
4 answers
Finding the limit of the sequence $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt n+1)$ [duplicate]
2 answers
Q) Let, $a_n ;=; left ( 1-frac1sqrt2 right ) ... left ( 1- frac1sqrtn+1 right )$ , $n geq 1$. Then $lim_nrightarrow infty a_n$
(A) equals $1$
(B) does not exist
(C) equals $frac1sqrtpi $
(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 :-
$left ( 1-frac1sqrt1+1 right ) * left ( 1-frac1sqrt2+1 right )*left ( 1-frac1sqrt3+1 right )*left ( 1-frac1sqrt4+1 right )*left ( 1-frac1sqrt5+1 right )*left ( 1-frac1sqrt6+1 right )*left ( 1-frac1sqrt7+1 right )*left ( 1-frac1sqrt8+1 right )*.........*left ( 1-frac1sqrtn+1 right )$
=$(0.293)*(0.423)*(0.5)*(0.553)*(0.622)*(0.647)*(0.667)* ....$
=0.009*...
So, here value is tending to zero. I think option $(D)$ is correct.
I have tried like this
$left ( fracsqrt2-1sqrt2 right )*left ( fracsqrt3-1sqrt3 right )*left ( fracsqrt4-1sqrt4 right )*.......left ( fracsqrt(n+1)-1sqrtn+1 right )$
= $left ( frac(sqrt2-1)*(sqrt3-1)*(sqrt4-1)*.......*(sqrtn+1-1)sqrt(n+1)! right )$
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
edited Feb 24 at 9:30
Michael Rozenberg
108k1895200
asked Feb 24 at 9:08
ankitankit
987
$endgroup$
marked as duplicate by Sil, Saucy O'Path, RRL, Lord Shark the Unknown, Pedro Feb 24 at 11:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
How can I find $lim_nto infty a_n$ [duplicate]
2 answers
How do I evaluate $prod_r=1^infty left (1-frac1sqrt r+1right)$?
2 answers
If $a_n=left(1-frac1sqrt2right)ldotsleft(1-frac1sqrtn+1right)$ then $lim_ntoinftya_n=?$ [duplicate]
4 answers
Finding the limit of the sequence $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt n+1)$ [duplicate]
2 answers
Q) Let, $a_n ;=; left ( 1-frac1sqrt2 right ) ... left ( 1- frac1sqrtn+1 right )$ , $n geq 1$. Then $lim_nrightarrow infty a_n$
(A) equals $1$
(B) does not exist
(C) equals $frac1sqrtpi $
(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 :-
$left ( 1-frac1sqrt1+1 right ) * left ( 1-frac1sqrt2+1 right )*left ( 1-frac1sqrt3+1 right )*left ( 1-frac1sqrt4+1 right )*left ( 1-frac1sqrt5+1 right )*left ( 1-frac1sqrt6+1 right )*left ( 1-frac1sqrt7+1 right )*left ( 1-frac1sqrt8+1 right )*.........*left ( 1-frac1sqrtn+1 right )$
=$(0.293)*(0.423)*(0.5)*(0.553)*(0.622)*(0.647)*(0.667)* ....$
=0.009*...
So, here value is tending to zero. I think option $(D)$ is correct.
I have tried like this
$left ( fracsqrt2-1sqrt2 right )*left ( fracsqrt3-1sqrt3 right )*left ( fracsqrt4-1sqrt4 right )*.......left ( fracsqrt(n+1)-1sqrtn+1 right )$
= $left ( frac(sqrt2-1)*(sqrt3-1)*(sqrt4-1)*.......*(sqrtn+1-1)sqrt(n+1)! right )$
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
edited Feb 24 at 9:30
Michael Rozenberg
108k1895200
asked Feb 24 at 9:08
ankitankit
987
$endgroup$
marked as duplicate by Sil, Saucy O'Path, RRL, Lord Shark the Unknown, Pedro Feb 24 at 11:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Using the inequality $1+x leq e^x$ which is true for all $x in mathbbR$, we have $$ a_n leq expleft -sum_k=2^n frac1sqrtk right leq C expleft -int_0^n fracmathrmdxsqrtx right = C e^-2sqrtn $$ 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 $a_n = e^-2sqrtn + mathcalO(1)$ as $ntoinfty$.
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– Sangchul Lee
Feb 24 at 9:12
1
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What would you do if it were $$left(1-frac12right)left(1-frac13right)cdotsleft(1-frac1n+1right)?$$
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– Lord Shark the Unknown
Feb 24 at 9:13
1
$begingroup$
Been asked here couple of times: $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt n+1)$, $mathbfa_n = (1-frac1sqrt2)...(1-frac1sqrtn+1)$, $n ge 1$ $Rightarrow$ $lim_nto infty a_n$ =?, Find the value of $lim_n rightarrow infty Big( 1-frac1sqrt 2 Big) cdots Big(1-frac1sqrt n+1 Big)$,
$endgroup$
– Sil
Feb 24 at 9:32
1
$begingroup$
If $a_n=left(1-frac1sqrt2right)ldotsleft(1-frac1sqrtn+1right)$ then $lim_ntoinftya_n=?$, How can I find $lim_nto infty a_n$ and Let $a_n=(1-frac1sqrt2)ldots(1-frac1sqrtn+1), n geq 1 $. Then $lim_limitsn rightarrow infty a_n $...
$endgroup$
– Sil
Feb 24 at 9:32
add a comment |
$begingroup$
This question already has an answer here:
How can I find $lim_nto infty a_n$ [duplicate]
2 answers
How do I evaluate $prod_r=1^infty left (1-frac1sqrt r+1right)$?
2 answers
If $a_n=left(1-frac1sqrt2right)ldotsleft(1-frac1sqrtn+1right)$ then $lim_ntoinftya_n=?$ [duplicate]
4 answers
Finding the limit of the sequence $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt n+1)$ [duplicate]
2 answers
Q) Let, $a_n ;=; left ( 1-frac1sqrt2 right ) ... left ( 1- frac1sqrtn+1 right )$ , $n geq 1$. Then $lim_nrightarrow infty a_n$
(A) equals $1$
(B) does not exist
(C) equals $frac1sqrtpi $
(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 :-
$left ( 1-frac1sqrt1+1 right ) * left ( 1-frac1sqrt2+1 right )*left ( 1-frac1sqrt3+1 right )*left ( 1-frac1sqrt4+1 right )*left ( 1-frac1sqrt5+1 right )*left ( 1-frac1sqrt6+1 right )*left ( 1-frac1sqrt7+1 right )*left ( 1-frac1sqrt8+1 right )*.........*left ( 1-frac1sqrtn+1 right )$
=$(0.293)*(0.423)*(0.5)*(0.553)*(0.622)*(0.647)*(0.667)* ....$
=0.009*...
So, here value is tending to zero. I think option $(D)$ is correct.
I have tried like this
$left ( fracsqrt2-1sqrt2 right )*left ( fracsqrt3-1sqrt3 right )*left ( fracsqrt4-1sqrt4 right )*.......left ( fracsqrt(n+1)-1sqrtn+1 right )$
= $left ( frac(sqrt2-1)*(sqrt3-1)*(sqrt4-1)*.......*(sqrtn+1-1)sqrt(n+1)! right )$
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
edited Feb 24 at 9:30
Michael Rozenberg
108k1895200
asked Feb 24 at 9:08
ankitankit
987
$endgroup$
This question already has an answer here:
How can I find $lim_nto infty a_n$ [duplicate]
2 answers
How do I evaluate $prod_r=1^infty left (1-frac1sqrt r+1right)$?
2 answers
If $a_n=left(1-frac1sqrt2right)ldotsleft(1-frac1sqrtn+1right)$ then $lim_ntoinftya_n=?$ [duplicate]
4 answers
Finding the limit of the sequence $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt n+1)$ [duplicate]
2 answers
Q) Let, $a_n ;=; left ( 1-frac1sqrt2 right ) ... left ( 1- frac1sqrtn+1 right )$ , $n geq 1$. Then $lim_nrightarrow infty a_n$
(A) equals $1$
(B) does not exist
(C) equals $frac1sqrtpi $
(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 :-
$left ( 1-frac1sqrt1+1 right ) * left ( 1-frac1sqrt2+1 right )*left ( 1-frac1sqrt3+1 right )*left ( 1-frac1sqrt4+1 right )*left ( 1-frac1sqrt5+1 right )*left ( 1-frac1sqrt6+1 right )*left ( 1-frac1sqrt7+1 right )*left ( 1-frac1sqrt8+1 right )*.........*left ( 1-frac1sqrtn+1 right )$
=$(0.293)*(0.423)*(0.5)*(0.553)*(0.622)*(0.647)*(0.667)* ....$
=0.009*...
So, here value is tending to zero. I think option $(D)$ is correct.
I have tried like this
$left ( fracsqrt2-1sqrt2 right )*left ( fracsqrt3-1sqrt3 right )*left ( fracsqrt4-1sqrt4 right )*.......left ( fracsqrt(n+1)-1sqrtn+1 right )$
= $left ( frac(sqrt2-1)*(sqrt3-1)*(sqrt4-1)*.......*(sqrtn+1-1)sqrt(n+1)! right )$
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.
This question already has an answer here:
How can I find $lim_nto infty a_n$ [duplicate]
2 answers
How do I evaluate $prod_r=1^infty left (1-frac1sqrt r+1right)$?
2 answers
If $a_n=left(1-frac1sqrt2right)ldotsleft(1-frac1sqrtn+1right)$ then $lim_ntoinftya_n=?$ [duplicate]
4 answers
Finding the limit of the sequence $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt n+1)$ [duplicate]
2 answers
calculus sequences-and-series limits products
calculus sequences-and-series limits products
edited Feb 24 at 9:30
Michael Rozenberg
108k1895200
asked Feb 24 at 9:08
ankitankit
987
edited Feb 24 at 9:30
Michael Rozenberg
108k1895200
asked Feb 24 at 9:08
ankitankit
987
edited Feb 24 at 9:30
Michael Rozenberg
108k1895200
edited Feb 24 at 9:30
Michael Rozenberg
108k1895200
edited Feb 24 at 9:30
Michael Rozenberg
108k1895200
108k1895200
asked Feb 24 at 9:08
ankitankit
987
asked Feb 24 at 9:08
ankitankit
987
asked Feb 24 at 9:08
ankitankit
987
987
marked as duplicate by Sil, Saucy O'Path, RRL, Lord Shark the Unknown, Pedro Feb 24 at 11:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Sil, Saucy O'Path, RRL, Lord Shark the Unknown, Pedro Feb 24 at 11:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Using the inequality $1+x leq e^x$ which is true for all $x in mathbbR$, we have $$ a_n leq expleft -sum_k=2^n frac1sqrtk right leq C expleft -int_0^n fracmathrmdxsqrtx right = C e^-2sqrtn $$ 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 $a_n = e^-2sqrtn + mathcalO(1)$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Feb 24 at 9:12
1
$begingroup$
What would you do if it were $$left(1-frac12right)left(1-frac13right)cdotsleft(1-frac1n+1right)?$$
$endgroup$
– Lord Shark the Unknown
Feb 24 at 9:13
1
$begingroup$
Been asked here couple of times: $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt n+1)$, $mathbfa_n = (1-frac1sqrt2)...(1-frac1sqrtn+1)$, $n ge 1$ $Rightarrow$ $lim_nto infty a_n$ =?, Find the value of $lim_n rightarrow infty Big( 1-frac1sqrt 2 Big) cdots Big(1-frac1sqrt n+1 Big)$,
$endgroup$
– Sil
Feb 24 at 9:32
1
$begingroup$
If $a_n=left(1-frac1sqrt2right)ldotsleft(1-frac1sqrtn+1right)$ then $lim_ntoinftya_n=?$, How can I find $lim_nto infty a_n$ and Let $a_n=(1-frac1sqrt2)ldots(1-frac1sqrtn+1), n geq 1 $. Then $lim_limitsn rightarrow infty a_n $...
$endgroup$
– Sil
Feb 24 at 9:32
add a comment |
$begingroup$
Using the inequality $1+x leq e^x$ which is true for all $x in mathbbR$, we have $$ a_n leq expleft -sum_k=2^n frac1sqrtk right leq C expleft -int_0^n fracmathrmdxsqrtx right = C e^-2sqrtn $$ 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 $a_n = e^-2sqrtn + mathcalO(1)$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Feb 24 at 9:12
1
$begingroup$
What would you do if it were $$left(1-frac12right)left(1-frac13right)cdotsleft(1-frac1n+1right)?$$
$endgroup$
– Lord Shark the Unknown
Feb 24 at 9:13
1
$begingroup$
Been asked here couple of times: $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt n+1)$, $mathbfa_n = (1-frac1sqrt2)...(1-frac1sqrtn+1)$, $n ge 1$ $Rightarrow$ $lim_nto infty a_n$ =?, Find the value of $lim_n rightarrow infty Big( 1-frac1sqrt 2 Big) cdots Big(1-frac1sqrt n+1 Big)$,
$endgroup$
– Sil
Feb 24 at 9:32
1
$begingroup$
If $a_n=left(1-frac1sqrt2right)ldotsleft(1-frac1sqrtn+1right)$ then $lim_ntoinftya_n=?$, How can I find $lim_nto infty a_n$ and Let $a_n=(1-frac1sqrt2)ldots(1-frac1sqrtn+1), n geq 1 $. Then $lim_limitsn rightarrow infty a_n $...
$endgroup$
– Sil
Feb 24 at 9:32
$begingroup$
Using the inequality $1+x leq e^x$ which is true for all $x in mathbbR$, we have $$ a_n leq expleft -sum_k=2^n frac1sqrtk right leq C expleft -int_0^n fracmathrmdxsqrtx right = C e^-2sqrtn $$ 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 $a_n = e^-2sqrtn + mathcalO(1)$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Feb 24 at 9:12
$begingroup$
Using the inequality $1+x leq e^x$ which is true for all $x in mathbbR$, we have $$ a_n leq expleft -sum_k=2^n frac1sqrtk right leq C expleft -int_0^n fracmathrmdxsqrtx right = C e^-2sqrtn $$ 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 $a_n = e^-2sqrtn + mathcalO(1)$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Feb 24 at 9:12
1
1
$begingroup$
What would you do if it were $$left(1-frac12right)left(1-frac13right)cdotsleft(1-frac1n+1right)?$$
$endgroup$
– Lord Shark the Unknown
Feb 24 at 9:13
$begingroup$
What would you do if it were $$left(1-frac12right)left(1-frac13right)cdotsleft(1-frac1n+1right)?$$
$endgroup$
– Lord Shark the Unknown
Feb 24 at 9:13
1
1
$begingroup$
Been asked here couple of times: $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt n+1)$, $mathbfa_n = (1-frac1sqrt2)...(1-frac1sqrtn+1)$, $n ge 1$ $Rightarrow$ $lim_nto infty a_n$ =?, Find the value of $lim_n rightarrow infty Big( 1-frac1sqrt 2 Big) cdots Big(1-frac1sqrt n+1 Big)$,
$endgroup$
– Sil
Feb 24 at 9:32
$begingroup$
Been asked here couple of times: $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt n+1)$, $mathbfa_n = (1-frac1sqrt2)...(1-frac1sqrtn+1)$, $n ge 1$ $Rightarrow$ $lim_nto infty a_n$ =?, Find the value of $lim_n rightarrow infty Big( 1-frac1sqrt 2 Big) cdots Big(1-frac1sqrt n+1 Big)$,
$endgroup$
– Sil
Feb 24 at 9:32
1
1
$begingroup$
If $a_n=left(1-frac1sqrt2right)ldotsleft(1-frac1sqrtn+1right)$ then $lim_ntoinftya_n=?$, How can I find $lim_nto infty a_n$ and Let $a_n=(1-frac1sqrt2)ldots(1-frac1sqrtn+1), n geq 1 $. Then $lim_limitsn rightarrow infty a_n $...
$endgroup$
– Sil
Feb 24 at 9:32
$begingroup$
If $a_n=left(1-frac1sqrt2right)ldotsleft(1-frac1sqrtn+1right)$ then $lim_ntoinftya_n=?$, How can I find $lim_nto infty a_n$ and Let $a_n=(1-frac1sqrt2)ldots(1-frac1sqrtn+1), n geq 1 $. Then $lim_limitsn rightarrow infty a_n $...
$endgroup$
– Sil
Feb 24 at 9:32
add a comment |
3 Answers
3
active
oldest
votes
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The hint:
$$0<a_n=prod_k=1^nleft(1-frac1sqrtk+1right)<prod_k=1^nleft(1-frac1k+1right)=frac1n+1rightarrow0$$
answered Feb 24 at 9:17
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
Michael.In 2 lines! :)
$endgroup$
– Peter Szilas
Feb 24 at 9:28
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 $b_n:=left(1+frac1sqrt 2right)cdots left(1+frac1sqrt n+1right)$, note that the product of corresponding factors is $left(1-frac1sqrt kright)left(1+frac1sqrt kright)=1-frac1k<1$, hence $a_nb_n<1$. On the other hand, by expanding the product and dropping lots of positive terms
$$b_nge 1+frac1sqrt 2+frac1sqrt 3+ldots+frac1sqrt n+1 $$
so that $b_n$ gets arbitrarily large. We conclude that $a_n$ gets arbitrarily small positive. In other words $a_nto 0$.
answered Feb 24 at 9:17
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Hagen.Very nice.
$endgroup$
– Peter Szilas
Feb 24 at 9:27
add a comment |
$begingroup$
Here is a simple approach to rule out the wrong options and therefore find the correct one.
I stress that this is not a proof that the limit is what it is, but a quick way of reasoning your way through a multiple choice question.
The numbers $1-1/sqrtk$ are all in $(0,1)$, so your sequence is positive and strictly decreasing.
Therefore it has a limit.
You have already computed enough terms to rule out the limits $1$ and $1/sqrtpi$.
Remember that it's decreasing.
The only remaining option is that the limit is zero.
answered Feb 24 at 12:14
Joonas IlmavirtaJoonas Ilmavirta
20.7k94282
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The hint:
$$0<a_n=prod_k=1^nleft(1-frac1sqrtk+1right)<prod_k=1^nleft(1-frac1k+1right)=frac1n+1rightarrow0$$
answered Feb 24 at 9:17
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
Michael.In 2 lines! :)
$endgroup$
– Peter Szilas
Feb 24 at 9:28
add a comment |
$begingroup$
The hint:
$$0<a_n=prod_k=1^nleft(1-frac1sqrtk+1right)<prod_k=1^nleft(1-frac1k+1right)=frac1n+1rightarrow0$$
answered Feb 24 at 9:17
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
Michael.In 2 lines! :)
$endgroup$
– Peter Szilas
Feb 24 at 9:28
add a comment |
$begingroup$
The hint:
$$0<a_n=prod_k=1^nleft(1-frac1sqrtk+1right)<prod_k=1^nleft(1-frac1k+1right)=frac1n+1rightarrow0$$
answered Feb 24 at 9:17
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
The hint:
$$0<a_n=prod_k=1^nleft(1-frac1sqrtk+1right)<prod_k=1^nleft(1-frac1k+1right)=frac1n+1rightarrow0$$
answered Feb 24 at 9:17
Michael RozenbergMichael Rozenberg
108k1895200
answered Feb 24 at 9:17
Michael RozenbergMichael Rozenberg
108k1895200
answered Feb 24 at 9:17
Michael RozenbergMichael Rozenberg
108k1895200
answered Feb 24 at 9:17
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
$begingroup$
Michael.In 2 lines! :)
$endgroup$
– Peter Szilas
Feb 24 at 9:28
add a comment |
$begingroup$
Michael.In 2 lines! :)
$endgroup$
– Peter Szilas
Feb 24 at 9:28
$begingroup$
Michael.In 2 lines! :)
$endgroup$
– Peter Szilas
Feb 24 at 9:28
$begingroup$
Michael.In 2 lines! :)
$endgroup$
– Peter Szilas
Feb 24 at 9:28
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 $b_n:=left(1+frac1sqrt 2right)cdots left(1+frac1sqrt n+1right)$, note that the product of corresponding factors is $left(1-frac1sqrt kright)left(1+frac1sqrt kright)=1-frac1k<1$, hence $a_nb_n<1$. On the other hand, by expanding the product and dropping lots of positive terms
$$b_nge 1+frac1sqrt 2+frac1sqrt 3+ldots+frac1sqrt n+1 $$
so that $b_n$ gets arbitrarily large. We conclude that $a_n$ gets arbitrarily small positive. In other words $a_nto 0$.
answered Feb 24 at 9:17
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Hagen.Very nice.
$endgroup$
– Peter Szilas
Feb 24 at 9:27
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 $b_n:=left(1+frac1sqrt 2right)cdots left(1+frac1sqrt n+1right)$, note that the product of corresponding factors is $left(1-frac1sqrt kright)left(1+frac1sqrt kright)=1-frac1k<1$, hence $a_nb_n<1$. On the other hand, by expanding the product and dropping lots of positive terms
$$b_nge 1+frac1sqrt 2+frac1sqrt 3+ldots+frac1sqrt n+1 $$
so that $b_n$ gets arbitrarily large. We conclude that $a_n$ gets arbitrarily small positive. In other words $a_nto 0$.
answered Feb 24 at 9:17
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Hagen.Very nice.
$endgroup$
– Peter Szilas
Feb 24 at 9:27
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 $b_n:=left(1+frac1sqrt 2right)cdots left(1+frac1sqrt n+1right)$, note that the product of corresponding factors is $left(1-frac1sqrt kright)left(1+frac1sqrt kright)=1-frac1k<1$, hence $a_nb_n<1$. On the other hand, by expanding the product and dropping lots of positive terms
$$b_nge 1+frac1sqrt 2+frac1sqrt 3+ldots+frac1sqrt n+1 $$
so that $b_n$ gets arbitrarily large. We conclude that $a_n$ gets arbitrarily small positive. In other words $a_nto 0$.
answered Feb 24 at 9:17
Hagen von EitzenHagen von Eitzen
283k23272507
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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 $b_n:=left(1+frac1sqrt 2right)cdots left(1+frac1sqrt n+1right)$, note that the product of corresponding factors is $left(1-frac1sqrt kright)left(1+frac1sqrt kright)=1-frac1k<1$, hence $a_nb_n<1$. On the other hand, by expanding the product and dropping lots of positive terms
$$b_nge 1+frac1sqrt 2+frac1sqrt 3+ldots+frac1sqrt n+1 $$
so that $b_n$ gets arbitrarily large. We conclude that $a_n$ gets arbitrarily small positive. In other words $a_nto 0$.
answered Feb 24 at 9:17
Hagen von EitzenHagen von Eitzen
283k23272507
answered Feb 24 at 9:17
Hagen von EitzenHagen von Eitzen
283k23272507
answered Feb 24 at 9:17
Hagen von EitzenHagen von Eitzen
283k23272507
answered Feb 24 at 9:17
Hagen von EitzenHagen von Eitzen
283k23272507
283k23272507
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Hagen.Very nice.
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– Peter Szilas
Feb 24 at 9:27
add a comment |
$begingroup$
Hagen.Very nice.
$endgroup$
– Peter Szilas
Feb 24 at 9:27
$begingroup$
Hagen.Very nice.
$endgroup$
– Peter Szilas
Feb 24 at 9:27
$begingroup$
Hagen.Very nice.
$endgroup$
– Peter Szilas
Feb 24 at 9:27
add a comment |
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Here is a simple approach to rule out the wrong options and therefore find the correct one.
I stress that this is not a proof that the limit is what it is, but a quick way of reasoning your way through a multiple choice question.
The numbers $1-1/sqrtk$ are all in $(0,1)$, so your sequence is positive and strictly decreasing.
Therefore it has a limit.
You have already computed enough terms to rule out the limits $1$ and $1/sqrtpi$.
Remember that it's decreasing.
The only remaining option is that the limit is zero.
answered Feb 24 at 12:14
Joonas IlmavirtaJoonas Ilmavirta
20.7k94282
$endgroup$
add a comment |
$begingroup$
Here is a simple approach to rule out the wrong options and therefore find the correct one.
I stress that this is not a proof that the limit is what it is, but a quick way of reasoning your way through a multiple choice question.
The numbers $1-1/sqrtk$ are all in $(0,1)$, so your sequence is positive and strictly decreasing.
Therefore it has a limit.
You have already computed enough terms to rule out the limits $1$ and $1/sqrtpi$.
Remember that it's decreasing.
The only remaining option is that the limit is zero.
answered Feb 24 at 12:14
Joonas IlmavirtaJoonas Ilmavirta
20.7k94282
$endgroup$
add a comment |
$begingroup$
Here is a simple approach to rule out the wrong options and therefore find the correct one.
I stress that this is not a proof that the limit is what it is, but a quick way of reasoning your way through a multiple choice question.
The numbers $1-1/sqrtk$ are all in $(0,1)$, so your sequence is positive and strictly decreasing.
Therefore it has a limit.
You have already computed enough terms to rule out the limits $1$ and $1/sqrtpi$.
Remember that it's decreasing.
The only remaining option is that the limit is zero.
answered Feb 24 at 12:14
Joonas IlmavirtaJoonas Ilmavirta
20.7k94282
$endgroup$
Here is a simple approach to rule out the wrong options and therefore find the correct one.
I stress that this is not a proof that the limit is what it is, but a quick way of reasoning your way through a multiple choice question.
The numbers $1-1/sqrtk$ are all in $(0,1)$, so your sequence is positive and strictly decreasing.
Therefore it has a limit.
You have already computed enough terms to rule out the limits $1$ and $1/sqrtpi$.
Remember that it's decreasing.
The only remaining option is that the limit is zero.
answered Feb 24 at 12:14
Joonas IlmavirtaJoonas Ilmavirta
20.7k94282
answered Feb 24 at 12:14
Joonas IlmavirtaJoonas Ilmavirta
20.7k94282
answered Feb 24 at 12:14
Joonas IlmavirtaJoonas Ilmavirta
20.7k94282
answered Feb 24 at 12:14
Joonas IlmavirtaJoonas Ilmavirta
20.7k94282
20.7k94282
add a comment |
add a comment |
$begingroup$
Using the inequality $1+x leq e^x$ which is true for all $x in mathbbR$, we have $$ a_n leq expleft -sum_k=2^n frac1sqrtk right leq C expleft -int_0^n fracmathrmdxsqrtx right = C e^-2sqrtn $$ 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 $a_n = e^-2sqrtn + mathcalO(1)$ as $ntoinfty$.
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– Sangchul Lee
Feb 24 at 9:12
1
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What would you do if it were $$left(1-frac12right)left(1-frac13right)cdotsleft(1-frac1n+1right)?$$
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– Lord Shark the Unknown
Feb 24 at 9:13
1
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Been asked here couple of times: $(1 - 1/sqrt 2) dotsm ( 1 - 1/sqrt n+1)$, $mathbfa_n = (1-frac1sqrt2)...(1-frac1sqrtn+1)$, $n ge 1$ $Rightarrow$ $lim_nto infty a_n$ =?, Find the value of $lim_n rightarrow infty Big( 1-frac1sqrt 2 Big) cdots Big(1-frac1sqrt n+1 Big)$,
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– Sil
Feb 24 at 9:32
1
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If $a_n=left(1-frac1sqrt2right)ldotsleft(1-frac1sqrtn+1right)$ then $lim_ntoinftya_n=?$, How can I find $lim_nto infty a_n$ and Let $a_n=(1-frac1sqrt2)ldots(1-frac1sqrtn+1), n geq 1 $. Then $lim_limitsn rightarrow infty a_n $...
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– Sil
Feb 24 at 9:32