Find the limit of $lim _ nrightarrow infty sum _ k=1 ^ n frac sqrt n ^ 2 - k ^ 2 n ^ 2 $ [closed]
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Need help to find the limit of $$lim _ nrightarrow infty sum _ k=1 ^ n frac sqrt n ^ 2 - k ^ 2 n ^ 2 $$
calculus sequences-and-series limits
edited Sep 1 at 9:38
Mattos
2,66321121
asked Sep 1 at 9:33
Sm1
91
closed as off-topic by Jam, Henrik, Claude Leibovici, Nosrati, Martin R Sep 1 at 11:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jam, Henrik, Claude Leibovici, Nosrati
 |Â
show 2 more comments
up vote
-2
down vote
favorite
Need help to find the limit of $$lim _ nrightarrow infty sum _ k=1 ^ n frac sqrt n ^ 2 - k ^ 2 n ^ 2 $$
calculus sequences-and-series limits
edited Sep 1 at 9:38
Mattos
2,66321121
asked Sep 1 at 9:33
Sm1
91
closed as off-topic by Jam, Henrik, Claude Leibovici, Nosrati, Martin R Sep 1 at 11:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jam, Henrik, Claude Leibovici, Nosrati
6
Hint: Riemann sum.
– Gabriel Romon
Sep 1 at 9:36
3
It's frowned upon to ask a question on this site without having shown your attempt at the problem. Update your post to include what you have tried.
– Mattos
Sep 1 at 9:39
Sry, new to the forum. Still have some difficulties with formula syntax.
– Sm1
Sep 1 at 9:45
read this math.meta.stackexchange.com/questions/5020/…
– Deepesh Meena
Sep 1 at 9:46
@Sm1 No worries - the reluctance of users to answer questions with little effort shown is mostly to stop people using the site to do their homework for them. Just try to make sure when you ask a question to show what you've tried. Here's a guide for formatting equations math.meta.stackexchange.com/questions/5020/…
– Jam
Sep 1 at 9:48
 |Â
show 2 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Need help to find the limit of $$lim _ nrightarrow infty sum _ k=1 ^ n frac sqrt n ^ 2 - k ^ 2 n ^ 2 $$
calculus sequences-and-series limits
edited Sep 1 at 9:38
Mattos
2,66321121
asked Sep 1 at 9:33
Sm1
91
Need help to find the limit of $$lim _ nrightarrow infty sum _ k=1 ^ n frac sqrt n ^ 2 - k ^ 2 n ^ 2 $$
calculus sequences-and-series limits
calculus sequences-and-series limits
edited Sep 1 at 9:38
Mattos
2,66321121
asked Sep 1 at 9:33
Sm1
91
edited Sep 1 at 9:38
Mattos
2,66321121
asked Sep 1 at 9:33
Sm1
91
edited Sep 1 at 9:38
Mattos
2,66321121
edited Sep 1 at 9:38
Mattos
2,66321121
edited Sep 1 at 9:38
Mattos
2,66321121
2,66321121
asked Sep 1 at 9:33
Sm1
91
asked Sep 1 at 9:33
Sm1
91
asked Sep 1 at 9:33
Sm1
91
91
closed as off-topic by Jam, Henrik, Claude Leibovici, Nosrati, Martin R Sep 1 at 11:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jam, Henrik, Claude Leibovici, Nosrati
closed as off-topic by Jam, Henrik, Claude Leibovici, Nosrati, Martin R Sep 1 at 11:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jam, Henrik, Claude Leibovici, Nosrati
6
Hint: Riemann sum.
– Gabriel Romon
Sep 1 at 9:36
3
It's frowned upon to ask a question on this site without having shown your attempt at the problem. Update your post to include what you have tried.
– Mattos
Sep 1 at 9:39
Sry, new to the forum. Still have some difficulties with formula syntax.
– Sm1
Sep 1 at 9:45
read this math.meta.stackexchange.com/questions/5020/…
– Deepesh Meena
Sep 1 at 9:46
@Sm1 No worries - the reluctance of users to answer questions with little effort shown is mostly to stop people using the site to do their homework for them. Just try to make sure when you ask a question to show what you've tried. Here's a guide for formatting equations math.meta.stackexchange.com/questions/5020/…
– Jam
Sep 1 at 9:48
 |Â
show 2 more comments
6
Hint: Riemann sum.
– Gabriel Romon
Sep 1 at 9:36
3
It's frowned upon to ask a question on this site without having shown your attempt at the problem. Update your post to include what you have tried.
– Mattos
Sep 1 at 9:39
Sry, new to the forum. Still have some difficulties with formula syntax.
– Sm1
Sep 1 at 9:45
read this math.meta.stackexchange.com/questions/5020/…
– Deepesh Meena
Sep 1 at 9:46
@Sm1 No worries - the reluctance of users to answer questions with little effort shown is mostly to stop people using the site to do their homework for them. Just try to make sure when you ask a question to show what you've tried. Here's a guide for formatting equations math.meta.stackexchange.com/questions/5020/…
– Jam
Sep 1 at 9:48
6
6
Hint: Riemann sum.
– Gabriel Romon
Sep 1 at 9:36
Hint: Riemann sum.
– Gabriel Romon
Sep 1 at 9:36
3
3
It's frowned upon to ask a question on this site without having shown your attempt at the problem. Update your post to include what you have tried.
– Mattos
Sep 1 at 9:39
It's frowned upon to ask a question on this site without having shown your attempt at the problem. Update your post to include what you have tried.
– Mattos
Sep 1 at 9:39
Sry, new to the forum. Still have some difficulties with formula syntax.
– Sm1
Sep 1 at 9:45
Sry, new to the forum. Still have some difficulties with formula syntax.
– Sm1
Sep 1 at 9:45
read this math.meta.stackexchange.com/questions/5020/…
– Deepesh Meena
Sep 1 at 9:46
read this math.meta.stackexchange.com/questions/5020/…
– Deepesh Meena
Sep 1 at 9:46
@Sm1 No worries - the reluctance of users to answer questions with little effort shown is mostly to stop people using the site to do their homework for them. Just try to make sure when you ask a question to show what you've tried. Here's a guide for formatting equations math.meta.stackexchange.com/questions/5020/…
– Jam
Sep 1 at 9:48
@Sm1 No worries - the reluctance of users to answer questions with little effort shown is mostly to stop people using the site to do their homework for them. Just try to make sure when you ask a question to show what you've tried. Here's a guide for formatting equations math.meta.stackexchange.com/questions/5020/…
– Jam
Sep 1 at 9:48
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Hint:
$$lim _ nrightarrow infty sum _ k=1 ^ n frac sqrt n ^ 2 - k ^ 2 n ^ 2 =lim_n to infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2$$
Replace $frac1n$ with $dx$ and $frackn$ with $x$
$$ lim _ nrightarrow infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2=int_0^1sqrt1-x^2dx$$
I obtained the bounds 0 and 1 using
$$int_a^bf(x),mathrmdx=lim_ntoinftysum_k^nfracb-anfleft(kfracb-anright)$$
answered Sep 1 at 9:56
Deepesh Meena
3,4982824
I'd like to note that you can justify this with $int_a^bf(x),mathrmdx=lim_ntoinftysum_k^nfracb-anfleft(kfracb-anright)$ (hence where the bounds $0$ and $1$ come from)
– Jam
Sep 1 at 10:09
Yes that's what I did. dodo you want me to include this in the answer?
– Deepesh Meena
Sep 1 at 10:10
You could do - the reason I pointed it out is that the bounds seem to come out of nowhere
– Jam
Sep 1 at 10:11
But one could also found the bounds like this $frackn$ is replaced by x but k can go max up to n thus x can go up to 1 also k is minimum 1 thus minimum x is 1/n but $n to infty$ thus minimum x is 0
– Deepesh Meena
Sep 1 at 10:15
add a comment |Â
up vote
0
down vote
As noticed by Riemann sum we obtain
$$lim _ nrightarrow infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2=int_0^1sqrt1-x^2dx=fracpi4$$
As an alternative, just to explore other ways, since for a fixed $k$ by binomial series we have
$$sqrt1-frack^2n^2=sum_j=0^infty binomfrac12jleft(-frack^2n^2right)^j $$
and therefore by Faulhaber's formula
$$sum _ k=1 ^ n frac1nsqrt1-frack^2n^2
=sum_j=0^infty left[binomfrac12jfrac(-1)^jn^2j+1sum_k=1^n k^2jright]
tosum_j=0^infty frac(-1)^jbinomfrac12j2j+1$$
and since for the binomial coefficient with $n=1/2$ we have
$$frac12choosej=2jchoosejfrac(-1)^j+12^2j(2j-1)implies binomfrac12jfrac(-1)^j2j+1=fracbinom2jj2^2j(1-4j^2)$$
we obtain that
$$sum_j=0^infty frac(-1)^jbinomfrac12j2j+1=sum_j=0^infty fracbinom2jj2^2j(1-4j^2)=fracpi4$$
answered Sep 1 at 10:31
gimusi
72k73888
How do you justify that last equality to $pi/4$?
– Jam
Sep 1 at 10:39
@Jam I have not anyidea about that! I'm assuming it is true from the result obtained by Riemann sum. I just wanted explore this other ways. Numerical extimation confirm the result. Do you have any idea how to derive the result in a different way without using Riemann sum?
– gimusi
Sep 1 at 10:42
If you format your links as[link name](link address), you can fit the whole URL in the link - otherwise I can't click on it.
– Jam
Sep 1 at 10:47
And I'm not sure about how to prove the equality to $pi/4$. I'd got as far as you but haven't thought of how to do the last step.
– Jam
Sep 1 at 10:47
@Jam sum 1
– gimusi
Sep 1 at 10:49
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint:
$$lim _ nrightarrow infty sum _ k=1 ^ n frac sqrt n ^ 2 - k ^ 2 n ^ 2 =lim_n to infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2$$
Replace $frac1n$ with $dx$ and $frackn$ with $x$
$$ lim _ nrightarrow infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2=int_0^1sqrt1-x^2dx$$
I obtained the bounds 0 and 1 using
$$int_a^bf(x),mathrmdx=lim_ntoinftysum_k^nfracb-anfleft(kfracb-anright)$$
answered Sep 1 at 9:56
Deepesh Meena
3,4982824
I'd like to note that you can justify this with $int_a^bf(x),mathrmdx=lim_ntoinftysum_k^nfracb-anfleft(kfracb-anright)$ (hence where the bounds $0$ and $1$ come from)
– Jam
Sep 1 at 10:09
Yes that's what I did. dodo you want me to include this in the answer?
– Deepesh Meena
Sep 1 at 10:10
You could do - the reason I pointed it out is that the bounds seem to come out of nowhere
– Jam
Sep 1 at 10:11
But one could also found the bounds like this $frackn$ is replaced by x but k can go max up to n thus x can go up to 1 also k is minimum 1 thus minimum x is 1/n but $n to infty$ thus minimum x is 0
– Deepesh Meena
Sep 1 at 10:15
add a comment |Â
up vote
2
down vote
accepted
Hint:
$$lim _ nrightarrow infty sum _ k=1 ^ n frac sqrt n ^ 2 - k ^ 2 n ^ 2 =lim_n to infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2$$
Replace $frac1n$ with $dx$ and $frackn$ with $x$
$$ lim _ nrightarrow infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2=int_0^1sqrt1-x^2dx$$
I obtained the bounds 0 and 1 using
$$int_a^bf(x),mathrmdx=lim_ntoinftysum_k^nfracb-anfleft(kfracb-anright)$$
answered Sep 1 at 9:56
Deepesh Meena
3,4982824
I'd like to note that you can justify this with $int_a^bf(x),mathrmdx=lim_ntoinftysum_k^nfracb-anfleft(kfracb-anright)$ (hence where the bounds $0$ and $1$ come from)
– Jam
Sep 1 at 10:09
Yes that's what I did. dodo you want me to include this in the answer?
– Deepesh Meena
Sep 1 at 10:10
You could do - the reason I pointed it out is that the bounds seem to come out of nowhere
– Jam
Sep 1 at 10:11
But one could also found the bounds like this $frackn$ is replaced by x but k can go max up to n thus x can go up to 1 also k is minimum 1 thus minimum x is 1/n but $n to infty$ thus minimum x is 0
– Deepesh Meena
Sep 1 at 10:15
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint:
$$lim _ nrightarrow infty sum _ k=1 ^ n frac sqrt n ^ 2 - k ^ 2 n ^ 2 =lim_n to infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2$$
Replace $frac1n$ with $dx$ and $frackn$ with $x$
$$ lim _ nrightarrow infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2=int_0^1sqrt1-x^2dx$$
I obtained the bounds 0 and 1 using
$$int_a^bf(x),mathrmdx=lim_ntoinftysum_k^nfracb-anfleft(kfracb-anright)$$
answered Sep 1 at 9:56
Deepesh Meena
3,4982824
Hint:
$$lim _ nrightarrow infty sum _ k=1 ^ n frac sqrt n ^ 2 - k ^ 2 n ^ 2 =lim_n to infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2$$
Replace $frac1n$ with $dx$ and $frackn$ with $x$
$$ lim _ nrightarrow infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2=int_0^1sqrt1-x^2dx$$
I obtained the bounds 0 and 1 using
$$int_a^bf(x),mathrmdx=lim_ntoinftysum_k^nfracb-anfleft(kfracb-anright)$$
answered Sep 1 at 9:56
Deepesh Meena
3,4982824
edited Sep 1 at 10:13
answered Sep 1 at 9:56
Deepesh Meena
3,4982824
answered Sep 1 at 9:56
Deepesh Meena
3,4982824
answered Sep 1 at 9:56
Deepesh Meena
3,4982824
3,4982824
I'd like to note that you can justify this with $int_a^bf(x),mathrmdx=lim_ntoinftysum_k^nfracb-anfleft(kfracb-anright)$ (hence where the bounds $0$ and $1$ come from)
– Jam
Sep 1 at 10:09
Yes that's what I did. dodo you want me to include this in the answer?
– Deepesh Meena
Sep 1 at 10:10
You could do - the reason I pointed it out is that the bounds seem to come out of nowhere
– Jam
Sep 1 at 10:11
But one could also found the bounds like this $frackn$ is replaced by x but k can go max up to n thus x can go up to 1 also k is minimum 1 thus minimum x is 1/n but $n to infty$ thus minimum x is 0
– Deepesh Meena
Sep 1 at 10:15
add a comment |Â
I'd like to note that you can justify this with $int_a^bf(x),mathrmdx=lim_ntoinftysum_k^nfracb-anfleft(kfracb-anright)$ (hence where the bounds $0$ and $1$ come from)
– Jam
Sep 1 at 10:09
Yes that's what I did. dodo you want me to include this in the answer?
– Deepesh Meena
Sep 1 at 10:10
You could do - the reason I pointed it out is that the bounds seem to come out of nowhere
– Jam
Sep 1 at 10:11
But one could also found the bounds like this $frackn$ is replaced by x but k can go max up to n thus x can go up to 1 also k is minimum 1 thus minimum x is 1/n but $n to infty$ thus minimum x is 0
– Deepesh Meena
Sep 1 at 10:15
I'd like to note that you can justify this with $int_a^bf(x),mathrmdx=lim_ntoinftysum_k^nfracb-anfleft(kfracb-anright)$ (hence where the bounds $0$ and $1$ come from)
– Jam
Sep 1 at 10:09
I'd like to note that you can justify this with $int_a^bf(x),mathrmdx=lim_ntoinftysum_k^nfracb-anfleft(kfracb-anright)$ (hence where the bounds $0$ and $1$ come from)
– Jam
Sep 1 at 10:09
Yes that's what I did. dodo you want me to include this in the answer?
– Deepesh Meena
Sep 1 at 10:10
Yes that's what I did. dodo you want me to include this in the answer?
– Deepesh Meena
Sep 1 at 10:10
You could do - the reason I pointed it out is that the bounds seem to come out of nowhere
– Jam
Sep 1 at 10:11
You could do - the reason I pointed it out is that the bounds seem to come out of nowhere
– Jam
Sep 1 at 10:11
But one could also found the bounds like this $frackn$ is replaced by x but k can go max up to n thus x can go up to 1 also k is minimum 1 thus minimum x is 1/n but $n to infty$ thus minimum x is 0
– Deepesh Meena
Sep 1 at 10:15
But one could also found the bounds like this $frackn$ is replaced by x but k can go max up to n thus x can go up to 1 also k is minimum 1 thus minimum x is 1/n but $n to infty$ thus minimum x is 0
– Deepesh Meena
Sep 1 at 10:15
add a comment |Â
up vote
0
down vote
As noticed by Riemann sum we obtain
$$lim _ nrightarrow infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2=int_0^1sqrt1-x^2dx=fracpi4$$
As an alternative, just to explore other ways, since for a fixed $k$ by binomial series we have
$$sqrt1-frack^2n^2=sum_j=0^infty binomfrac12jleft(-frack^2n^2right)^j $$
and therefore by Faulhaber's formula
$$sum _ k=1 ^ n frac1nsqrt1-frack^2n^2
=sum_j=0^infty left[binomfrac12jfrac(-1)^jn^2j+1sum_k=1^n k^2jright]
tosum_j=0^infty frac(-1)^jbinomfrac12j2j+1$$
and since for the binomial coefficient with $n=1/2$ we have
$$frac12choosej=2jchoosejfrac(-1)^j+12^2j(2j-1)implies binomfrac12jfrac(-1)^j2j+1=fracbinom2jj2^2j(1-4j^2)$$
we obtain that
$$sum_j=0^infty frac(-1)^jbinomfrac12j2j+1=sum_j=0^infty fracbinom2jj2^2j(1-4j^2)=fracpi4$$
answered Sep 1 at 10:31
gimusi
72k73888
How do you justify that last equality to $pi/4$?
– Jam
Sep 1 at 10:39
@Jam I have not anyidea about that! I'm assuming it is true from the result obtained by Riemann sum. I just wanted explore this other ways. Numerical extimation confirm the result. Do you have any idea how to derive the result in a different way without using Riemann sum?
– gimusi
Sep 1 at 10:42
If you format your links as[link name](link address), you can fit the whole URL in the link - otherwise I can't click on it.
– Jam
Sep 1 at 10:47
And I'm not sure about how to prove the equality to $pi/4$. I'd got as far as you but haven't thought of how to do the last step.
– Jam
Sep 1 at 10:47
@Jam sum 1
– gimusi
Sep 1 at 10:49
 |Â
show 4 more comments
up vote
0
down vote
As noticed by Riemann sum we obtain
$$lim _ nrightarrow infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2=int_0^1sqrt1-x^2dx=fracpi4$$
As an alternative, just to explore other ways, since for a fixed $k$ by binomial series we have
$$sqrt1-frack^2n^2=sum_j=0^infty binomfrac12jleft(-frack^2n^2right)^j $$
and therefore by Faulhaber's formula
$$sum _ k=1 ^ n frac1nsqrt1-frack^2n^2
=sum_j=0^infty left[binomfrac12jfrac(-1)^jn^2j+1sum_k=1^n k^2jright]
tosum_j=0^infty frac(-1)^jbinomfrac12j2j+1$$
and since for the binomial coefficient with $n=1/2$ we have
$$frac12choosej=2jchoosejfrac(-1)^j+12^2j(2j-1)implies binomfrac12jfrac(-1)^j2j+1=fracbinom2jj2^2j(1-4j^2)$$
we obtain that
$$sum_j=0^infty frac(-1)^jbinomfrac12j2j+1=sum_j=0^infty fracbinom2jj2^2j(1-4j^2)=fracpi4$$
answered Sep 1 at 10:31
gimusi
72k73888
How do you justify that last equality to $pi/4$?
– Jam
Sep 1 at 10:39
@Jam I have not anyidea about that! I'm assuming it is true from the result obtained by Riemann sum. I just wanted explore this other ways. Numerical extimation confirm the result. Do you have any idea how to derive the result in a different way without using Riemann sum?
– gimusi
Sep 1 at 10:42
If you format your links as[link name](link address), you can fit the whole URL in the link - otherwise I can't click on it.
– Jam
Sep 1 at 10:47
And I'm not sure about how to prove the equality to $pi/4$. I'd got as far as you but haven't thought of how to do the last step.
– Jam
Sep 1 at 10:47
@Jam sum 1
– gimusi
Sep 1 at 10:49
 |Â
show 4 more comments
up vote
0
down vote
up vote
0
down vote
As noticed by Riemann sum we obtain
$$lim _ nrightarrow infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2=int_0^1sqrt1-x^2dx=fracpi4$$
As an alternative, just to explore other ways, since for a fixed $k$ by binomial series we have
$$sqrt1-frack^2n^2=sum_j=0^infty binomfrac12jleft(-frack^2n^2right)^j $$
and therefore by Faulhaber's formula
$$sum _ k=1 ^ n frac1nsqrt1-frack^2n^2
=sum_j=0^infty left[binomfrac12jfrac(-1)^jn^2j+1sum_k=1^n k^2jright]
tosum_j=0^infty frac(-1)^jbinomfrac12j2j+1$$
and since for the binomial coefficient with $n=1/2$ we have
$$frac12choosej=2jchoosejfrac(-1)^j+12^2j(2j-1)implies binomfrac12jfrac(-1)^j2j+1=fracbinom2jj2^2j(1-4j^2)$$
we obtain that
$$sum_j=0^infty frac(-1)^jbinomfrac12j2j+1=sum_j=0^infty fracbinom2jj2^2j(1-4j^2)=fracpi4$$
answered Sep 1 at 10:31
gimusi
72k73888
As noticed by Riemann sum we obtain
$$lim _ nrightarrow infty sum _ k=1 ^ n frac1nsqrt1-frack^2n^2=int_0^1sqrt1-x^2dx=fracpi4$$
As an alternative, just to explore other ways, since for a fixed $k$ by binomial series we have
$$sqrt1-frack^2n^2=sum_j=0^infty binomfrac12jleft(-frack^2n^2right)^j $$
and therefore by Faulhaber's formula
$$sum _ k=1 ^ n frac1nsqrt1-frack^2n^2
=sum_j=0^infty left[binomfrac12jfrac(-1)^jn^2j+1sum_k=1^n k^2jright]
tosum_j=0^infty frac(-1)^jbinomfrac12j2j+1$$
and since for the binomial coefficient with $n=1/2$ we have
$$frac12choosej=2jchoosejfrac(-1)^j+12^2j(2j-1)implies binomfrac12jfrac(-1)^j2j+1=fracbinom2jj2^2j(1-4j^2)$$
we obtain that
$$sum_j=0^infty frac(-1)^jbinomfrac12j2j+1=sum_j=0^infty fracbinom2jj2^2j(1-4j^2)=fracpi4$$
answered Sep 1 at 10:31
gimusi
72k73888
edited Sep 1 at 11:26
answered Sep 1 at 10:31
gimusi
72k73888
answered Sep 1 at 10:31
gimusi
72k73888
answered Sep 1 at 10:31
gimusi
72k73888
72k73888
How do you justify that last equality to $pi/4$?
– Jam
Sep 1 at 10:39
@Jam I have not anyidea about that! I'm assuming it is true from the result obtained by Riemann sum. I just wanted explore this other ways. Numerical extimation confirm the result. Do you have any idea how to derive the result in a different way without using Riemann sum?
– gimusi
Sep 1 at 10:42
If you format your links as[link name](link address), you can fit the whole URL in the link - otherwise I can't click on it.
– Jam
Sep 1 at 10:47
And I'm not sure about how to prove the equality to $pi/4$. I'd got as far as you but haven't thought of how to do the last step.
– Jam
Sep 1 at 10:47
@Jam sum 1
– gimusi
Sep 1 at 10:49
 |Â
show 4 more comments
How do you justify that last equality to $pi/4$?
– Jam
Sep 1 at 10:39
@Jam I have not anyidea about that! I'm assuming it is true from the result obtained by Riemann sum. I just wanted explore this other ways. Numerical extimation confirm the result. Do you have any idea how to derive the result in a different way without using Riemann sum?
– gimusi
Sep 1 at 10:42
If you format your links as[link name](link address), you can fit the whole URL in the link - otherwise I can't click on it.
– Jam
Sep 1 at 10:47
And I'm not sure about how to prove the equality to $pi/4$. I'd got as far as you but haven't thought of how to do the last step.
– Jam
Sep 1 at 10:47
@Jam sum 1
– gimusi
Sep 1 at 10:49
How do you justify that last equality to $pi/4$?
– Jam
Sep 1 at 10:39
How do you justify that last equality to $pi/4$?
– Jam
Sep 1 at 10:39
@Jam I have not anyidea about that! I'm assuming it is true from the result obtained by Riemann sum. I just wanted explore this other ways. Numerical extimation confirm the result. Do you have any idea how to derive the result in a different way without using Riemann sum?
– gimusi
Sep 1 at 10:42
@Jam I have not anyidea about that! I'm assuming it is true from the result obtained by Riemann sum. I just wanted explore this other ways. Numerical extimation confirm the result. Do you have any idea how to derive the result in a different way without using Riemann sum?
– gimusi
Sep 1 at 10:42
If you format your links as
[link name](link address), you can fit the whole URL in the link - otherwise I can't click on it.– Jam
Sep 1 at 10:47
If you format your links as
[link name](link address), you can fit the whole URL in the link - otherwise I can't click on it.– Jam
Sep 1 at 10:47
And I'm not sure about how to prove the equality to $pi/4$. I'd got as far as you but haven't thought of how to do the last step.
– Jam
Sep 1 at 10:47
And I'm not sure about how to prove the equality to $pi/4$. I'd got as far as you but haven't thought of how to do the last step.
– Jam
Sep 1 at 10:47
@Jam sum 1
– gimusi
Sep 1 at 10:49
@Jam sum 1
– gimusi
Sep 1 at 10:49
 |Â
show 4 more comments
6
Hint: Riemann sum.
– Gabriel Romon
Sep 1 at 9:36
3
It's frowned upon to ask a question on this site without having shown your attempt at the problem. Update your post to include what you have tried.
– Mattos
Sep 1 at 9:39
Sry, new to the forum. Still have some difficulties with formula syntax.
– Sm1
Sep 1 at 9:45
read this math.meta.stackexchange.com/questions/5020/…
– Deepesh Meena
Sep 1 at 9:46
@Sm1 No worries - the reluctance of users to answer questions with little effort shown is mostly to stop people using the site to do their homework for them. Just try to make sure when you ask a question to show what you've tried. Here's a guide for formatting equations math.meta.stackexchange.com/questions/5020/…
– Jam
Sep 1 at 9:48