Calculate $sum_k=2^infty frac1k^2 - 1$
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
I'm wondering if someone could check on this working:
$$sum_k=2^infty frac1k^2 - 1 = sum_k=2^infty frac1(k - 1)(k + 1) = frac12sum_k=2^infty frac1k-1 - frac1k+1$$
This is a nice telescoping thing that has a $k$-th partial sum that looks like this:
$$frac11 - frac13 + frac12 - frac14 + frac13 - frac15 + ... + frac1k-1 - frac1k+1 = 1 + frac12 - frac1k+1 - frac1k+1$$
Skipping some formalities we find:
$$sum_k=2^infty frac1k^2 - 1 = frac12 . lim_k rightarrow inftyfrac32-frac2k+1 = frac12 . frac32 = frac34$$
real-analysis
asked Sep 8 at 6:04
Florian Suess
329110
add a comment |Â
up vote
3
down vote
favorite
I'm wondering if someone could check on this working:
$$sum_k=2^infty frac1k^2 - 1 = sum_k=2^infty frac1(k - 1)(k + 1) = frac12sum_k=2^infty frac1k-1 - frac1k+1$$
This is a nice telescoping thing that has a $k$-th partial sum that looks like this:
$$frac11 - frac13 + frac12 - frac14 + frac13 - frac15 + ... + frac1k-1 - frac1k+1 = 1 + frac12 - frac1k+1 - frac1k+1$$
Skipping some formalities we find:
$$sum_k=2^infty frac1k^2 - 1 = frac12 . lim_k rightarrow inftyfrac32-frac2k+1 = frac12 . frac32 = frac34$$
real-analysis
asked Sep 8 at 6:04
Florian Suess
329110
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm wondering if someone could check on this working:
$$sum_k=2^infty frac1k^2 - 1 = sum_k=2^infty frac1(k - 1)(k + 1) = frac12sum_k=2^infty frac1k-1 - frac1k+1$$
This is a nice telescoping thing that has a $k$-th partial sum that looks like this:
$$frac11 - frac13 + frac12 - frac14 + frac13 - frac15 + ... + frac1k-1 - frac1k+1 = 1 + frac12 - frac1k+1 - frac1k+1$$
Skipping some formalities we find:
$$sum_k=2^infty frac1k^2 - 1 = frac12 . lim_k rightarrow inftyfrac32-frac2k+1 = frac12 . frac32 = frac34$$
real-analysis
asked Sep 8 at 6:04
Florian Suess
329110
I'm wondering if someone could check on this working:
$$sum_k=2^infty frac1k^2 - 1 = sum_k=2^infty frac1(k - 1)(k + 1) = frac12sum_k=2^infty frac1k-1 - frac1k+1$$
This is a nice telescoping thing that has a $k$-th partial sum that looks like this:
$$frac11 - frac13 + frac12 - frac14 + frac13 - frac15 + ... + frac1k-1 - frac1k+1 = 1 + frac12 - frac1k+1 - frac1k+1$$
Skipping some formalities we find:
$$sum_k=2^infty frac1k^2 - 1 = frac12 . lim_k rightarrow inftyfrac32-frac2k+1 = frac12 . frac32 = frac34$$
real-analysis
real-analysis
asked Sep 8 at 6:04
Florian Suess
329110
asked Sep 8 at 6:04
Florian Suess
329110
asked Sep 8 at 6:04
Florian Suess
329110
asked Sep 8 at 6:04
Florian Suess
329110
asked Sep 8 at 6:04
Florian Suess
329110
329110
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
You have
$$
...= 1 + frac12 - frac1k+1 - frac1k+1
$$
and there should be
$$
= 1 + frac12 - frac1colorcyank - frac1k+1
$$
but it is easy to correct and even with this mistake does not change the result.
answered Sep 8 at 6:08
Przemysław Scherwentke
11.8k52751
1
Actually now reviewing it, I'm actually thinking it should be $1 + frac12 - frac1colorredk - frac1k+1$, I'm drawing out a few sequences and I find the remaining terms (with k in it) are $- frac1(k - 1) + 1 - frac1k+1$
– Florian Suess
Sep 8 at 6:13
Yeah it's always the last two negative terms.
– Florian Suess
Sep 8 at 6:14
Nevertheless, it doesn't change the result, but thanks for checking the other stuff!
– Florian Suess
Sep 8 at 6:15
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You have
$$
...= 1 + frac12 - frac1k+1 - frac1k+1
$$
and there should be
$$
= 1 + frac12 - frac1colorcyank - frac1k+1
$$
but it is easy to correct and even with this mistake does not change the result.
answered Sep 8 at 6:08
Przemysław Scherwentke
11.8k52751
1
Actually now reviewing it, I'm actually thinking it should be $1 + frac12 - frac1colorredk - frac1k+1$, I'm drawing out a few sequences and I find the remaining terms (with k in it) are $- frac1(k - 1) + 1 - frac1k+1$
– Florian Suess
Sep 8 at 6:13
Yeah it's always the last two negative terms.
– Florian Suess
Sep 8 at 6:14
Nevertheless, it doesn't change the result, but thanks for checking the other stuff!
– Florian Suess
Sep 8 at 6:15
add a comment |Â
up vote
2
down vote
accepted
You have
$$
...= 1 + frac12 - frac1k+1 - frac1k+1
$$
and there should be
$$
= 1 + frac12 - frac1colorcyank - frac1k+1
$$
but it is easy to correct and even with this mistake does not change the result.
answered Sep 8 at 6:08
Przemysław Scherwentke
11.8k52751
1
Actually now reviewing it, I'm actually thinking it should be $1 + frac12 - frac1colorredk - frac1k+1$, I'm drawing out a few sequences and I find the remaining terms (with k in it) are $- frac1(k - 1) + 1 - frac1k+1$
– Florian Suess
Sep 8 at 6:13
Yeah it's always the last two negative terms.
– Florian Suess
Sep 8 at 6:14
Nevertheless, it doesn't change the result, but thanks for checking the other stuff!
– Florian Suess
Sep 8 at 6:15
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You have
$$
...= 1 + frac12 - frac1k+1 - frac1k+1
$$
and there should be
$$
= 1 + frac12 - frac1colorcyank - frac1k+1
$$
but it is easy to correct and even with this mistake does not change the result.
answered Sep 8 at 6:08
Przemysław Scherwentke
11.8k52751
You have
$$
...= 1 + frac12 - frac1k+1 - frac1k+1
$$
and there should be
$$
= 1 + frac12 - frac1colorcyank - frac1k+1
$$
but it is easy to correct and even with this mistake does not change the result.
answered Sep 8 at 6:08
Przemysław Scherwentke
11.8k52751
edited Sep 8 at 6:16
answered Sep 8 at 6:08
Przemysław Scherwentke
11.8k52751
answered Sep 8 at 6:08
Przemysław Scherwentke
11.8k52751
answered Sep 8 at 6:08
Przemysław Scherwentke
11.8k52751
11.8k52751
1
Actually now reviewing it, I'm actually thinking it should be $1 + frac12 - frac1colorredk - frac1k+1$, I'm drawing out a few sequences and I find the remaining terms (with k in it) are $- frac1(k - 1) + 1 - frac1k+1$
– Florian Suess
Sep 8 at 6:13
Yeah it's always the last two negative terms.
– Florian Suess
Sep 8 at 6:14
Nevertheless, it doesn't change the result, but thanks for checking the other stuff!
– Florian Suess
Sep 8 at 6:15
add a comment |Â
1
Actually now reviewing it, I'm actually thinking it should be $1 + frac12 - frac1colorredk - frac1k+1$, I'm drawing out a few sequences and I find the remaining terms (with k in it) are $- frac1(k - 1) + 1 - frac1k+1$
– Florian Suess
Sep 8 at 6:13
Yeah it's always the last two negative terms.
– Florian Suess
Sep 8 at 6:14
Nevertheless, it doesn't change the result, but thanks for checking the other stuff!
– Florian Suess
Sep 8 at 6:15
1
1
Actually now reviewing it, I'm actually thinking it should be $1 + frac12 - frac1colorredk - frac1k+1$, I'm drawing out a few sequences and I find the remaining terms (with k in it) are $- frac1(k - 1) + 1 - frac1k+1$
– Florian Suess
Sep 8 at 6:13
Actually now reviewing it, I'm actually thinking it should be $1 + frac12 - frac1colorredk - frac1k+1$, I'm drawing out a few sequences and I find the remaining terms (with k in it) are $- frac1(k - 1) + 1 - frac1k+1$
– Florian Suess
Sep 8 at 6:13
Yeah it's always the last two negative terms.
– Florian Suess
Sep 8 at 6:14
Yeah it's always the last two negative terms.
– Florian Suess
Sep 8 at 6:14
Nevertheless, it doesn't change the result, but thanks for checking the other stuff!
– Florian Suess
Sep 8 at 6:15
Nevertheless, it doesn't change the result, but thanks for checking the other stuff!
– Florian Suess
Sep 8 at 6:15
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2909323%2fcalculate-sum-k-2-infty-frac1k2-1%23new-answer', 'question_page');
);
Post as a guest
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
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
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
