$1+frac {1}{4}(1+frac {1}{4}) +frac {1}{9}(1+frac {1}{4} +frac {1}{9})+…$
up vote
7
down vote
favorite
Show that $$1+frac {1}{4} bigg(1+frac {1}{4}bigg) +frac {1}{9} bigg(1+frac {1}{4} +frac {1}{9}bigg)+.....$$
converges.
Can you find the exact value of the sum.
My effort:
I have proved the convergence with comparing to $$bigg(sum _1^infty frac {1}{n^2}bigg)^2$$
I have not figure out the exact sum.
Any suggestions??
calculus sequences-and-series convergence
edited Nov 15 at 16:14
Rebellos
11.5k21040
asked Nov 15 at 16:11
Mohammad Riazi-Kermani
40.2k41958
add a comment |
up vote
7
down vote
favorite
Show that $$1+frac {1}{4} bigg(1+frac {1}{4}bigg) +frac {1}{9} bigg(1+frac {1}{4} +frac {1}{9}bigg)+.....$$
converges.
Can you find the exact value of the sum.
My effort:
I have proved the convergence with comparing to $$bigg(sum _1^infty frac {1}{n^2}bigg)^2$$
I have not figure out the exact sum.
Any suggestions??
calculus sequences-and-series convergence
edited Nov 15 at 16:14
Rebellos
11.5k21040
asked Nov 15 at 16:11
Mohammad Riazi-Kermani
40.2k41958
The sum is equal to $sum_nfrac{lceil d(n)/2rceil}{n^2}$, if this helps.
– ajotatxe
Nov 15 at 16:19
According to Mathematica, $sum _{n=1}^{infty } frac{sum _{i=1}^n frac{1}{i^2}}{n^2}={7pi^4over360}$. Maybe this gives you an idea of how to derive the sum.
– Steve Kass
Nov 15 at 16:26
@SteveKass It should be related to $zeta(4)$
– ajotatxe
Nov 15 at 16:28
@SteveKass Thanks for the comment, yes sometimes the answer helps to figure out a solution.
– Mohammad Riazi-Kermani
Nov 15 at 16:28
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Show that $$1+frac {1}{4} bigg(1+frac {1}{4}bigg) +frac {1}{9} bigg(1+frac {1}{4} +frac {1}{9}bigg)+.....$$
converges.
Can you find the exact value of the sum.
My effort:
I have proved the convergence with comparing to $$bigg(sum _1^infty frac {1}{n^2}bigg)^2$$
I have not figure out the exact sum.
Any suggestions??
calculus sequences-and-series convergence
edited Nov 15 at 16:14
Rebellos
11.5k21040
asked Nov 15 at 16:11
Mohammad Riazi-Kermani
40.2k41958
Show that $$1+frac {1}{4} bigg(1+frac {1}{4}bigg) +frac {1}{9} bigg(1+frac {1}{4} +frac {1}{9}bigg)+.....$$
converges.
Can you find the exact value of the sum.
My effort:
I have proved the convergence with comparing to $$bigg(sum _1^infty frac {1}{n^2}bigg)^2$$
I have not figure out the exact sum.
Any suggestions??
calculus sequences-and-series convergence
calculus sequences-and-series convergence
edited Nov 15 at 16:14
Rebellos
11.5k21040
asked Nov 15 at 16:11
Mohammad Riazi-Kermani
40.2k41958
edited Nov 15 at 16:14
Rebellos
11.5k21040
asked Nov 15 at 16:11
Mohammad Riazi-Kermani
40.2k41958
edited Nov 15 at 16:14
Rebellos
11.5k21040
edited Nov 15 at 16:14
Rebellos
11.5k21040
edited Nov 15 at 16:14
Rebellos
11.5k21040
11.5k21040
asked Nov 15 at 16:11
Mohammad Riazi-Kermani
40.2k41958
asked Nov 15 at 16:11
Mohammad Riazi-Kermani
40.2k41958
asked Nov 15 at 16:11
Mohammad Riazi-Kermani
40.2k41958
40.2k41958
The sum is equal to $sum_nfrac{lceil d(n)/2rceil}{n^2}$, if this helps.
– ajotatxe
Nov 15 at 16:19
According to Mathematica, $sum _{n=1}^{infty } frac{sum _{i=1}^n frac{1}{i^2}}{n^2}={7pi^4over360}$. Maybe this gives you an idea of how to derive the sum.
– Steve Kass
Nov 15 at 16:26
@SteveKass It should be related to $zeta(4)$
– ajotatxe
Nov 15 at 16:28
@SteveKass Thanks for the comment, yes sometimes the answer helps to figure out a solution.
– Mohammad Riazi-Kermani
Nov 15 at 16:28
add a comment |
The sum is equal to $sum_nfrac{lceil d(n)/2rceil}{n^2}$, if this helps.
– ajotatxe
Nov 15 at 16:19
According to Mathematica, $sum _{n=1}^{infty } frac{sum _{i=1}^n frac{1}{i^2}}{n^2}={7pi^4over360}$. Maybe this gives you an idea of how to derive the sum.
– Steve Kass
Nov 15 at 16:26
@SteveKass It should be related to $zeta(4)$
– ajotatxe
Nov 15 at 16:28
@SteveKass Thanks for the comment, yes sometimes the answer helps to figure out a solution.
– Mohammad Riazi-Kermani
Nov 15 at 16:28
The sum is equal to $sum_nfrac{lceil d(n)/2rceil}{n^2}$, if this helps.
– ajotatxe
Nov 15 at 16:19
The sum is equal to $sum_nfrac{lceil d(n)/2rceil}{n^2}$, if this helps.
– ajotatxe
Nov 15 at 16:19
According to Mathematica, $sum _{n=1}^{infty } frac{sum _{i=1}^n frac{1}{i^2}}{n^2}={7pi^4over360}$. Maybe this gives you an idea of how to derive the sum.
– Steve Kass
Nov 15 at 16:26
According to Mathematica, $sum _{n=1}^{infty } frac{sum _{i=1}^n frac{1}{i^2}}{n^2}={7pi^4over360}$. Maybe this gives you an idea of how to derive the sum.
– Steve Kass
Nov 15 at 16:26
@SteveKass It should be related to $zeta(4)$
– ajotatxe
Nov 15 at 16:28
@SteveKass It should be related to $zeta(4)$
– ajotatxe
Nov 15 at 16:28
@SteveKass Thanks for the comment, yes sometimes the answer helps to figure out a solution.
– Mohammad Riazi-Kermani
Nov 15 at 16:28
@SteveKass Thanks for the comment, yes sometimes the answer helps to figure out a solution.
– Mohammad Riazi-Kermani
Nov 15 at 16:28
add a comment |
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
$$
2S = sum_{ileq j} frac{1}{i^{2}j^{2}} + sum_{igeq j} frac{1}{i^{2}j^{2}} = left(sum_{ngeq 1}frac{1}{n^{2}}right)^{2} + sum_{ngeq 1}frac{1}{n^{4}} = frac{pi^{4}}{36} + frac{pi^{4}}{90}
$$
answered Nov 15 at 16:28
Seewoo Lee
5,756825
1
@Seewood Lee Good answer, thanks.
– Mohammad Riazi-Kermani
Nov 15 at 16:32
1
We can generalize the result to $$sum_{ile j}frac{1}{i^r j^r}=frac12left(zeta(r)^2+zeta(2r)right).$$
– Tianlalu
Nov 15 at 16:43
add a comment |
up vote
2
down vote
Suppose that $sum_{i=0}^infty a_i$ is an absolutely convergent series. Then (where $i$ and $j$ range over the nonnegative integers)
$$2sum_{i<=j} a_ia_j=\
2sum_{i<j} a_ia_j+2sum_{i=j} a_ia_j=\
sum_{i<j} a_ia_j+sum_{i>j} a_ia_j+color{green}{2sum_{i=j} a_ia_j}=\
left(sum_{i<j} a_ia_j+color{green}{sum_{i=j} a_ia_j}+sum_{i>j} a_ia_jright)+color{green}{sum_{i=j} a_ia_j}=\
color{red}{{sum_{i,j} a_ia_j}}+color{blue}{sum_{i=j} a_ia_j}=\
color{red}{left(sum_{i} a_iright)^2}+color{blue}{sum_{i} (a_i)^2}.$$
The question here is answered by this identity for $displaystyle a_i={1over i^2}$.
answered Nov 15 at 21:22
Steve Kass
10.8k11429
Very beautiful, thank for the answer..
– Mohammad Riazi-Kermani
Nov 15 at 21:33
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
$$
2S = sum_{ileq j} frac{1}{i^{2}j^{2}} + sum_{igeq j} frac{1}{i^{2}j^{2}} = left(sum_{ngeq 1}frac{1}{n^{2}}right)^{2} + sum_{ngeq 1}frac{1}{n^{4}} = frac{pi^{4}}{36} + frac{pi^{4}}{90}
$$
answered Nov 15 at 16:28
Seewoo Lee
5,756825
1
@Seewood Lee Good answer, thanks.
– Mohammad Riazi-Kermani
Nov 15 at 16:32
1
We can generalize the result to $$sum_{ile j}frac{1}{i^r j^r}=frac12left(zeta(r)^2+zeta(2r)right).$$
– Tianlalu
Nov 15 at 16:43
add a comment |
up vote
6
down vote
accepted
$$
2S = sum_{ileq j} frac{1}{i^{2}j^{2}} + sum_{igeq j} frac{1}{i^{2}j^{2}} = left(sum_{ngeq 1}frac{1}{n^{2}}right)^{2} + sum_{ngeq 1}frac{1}{n^{4}} = frac{pi^{4}}{36} + frac{pi^{4}}{90}
$$
answered Nov 15 at 16:28
Seewoo Lee
5,756825
1
@Seewood Lee Good answer, thanks.
– Mohammad Riazi-Kermani
Nov 15 at 16:32
1
We can generalize the result to $$sum_{ile j}frac{1}{i^r j^r}=frac12left(zeta(r)^2+zeta(2r)right).$$
– Tianlalu
Nov 15 at 16:43
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
$$
2S = sum_{ileq j} frac{1}{i^{2}j^{2}} + sum_{igeq j} frac{1}{i^{2}j^{2}} = left(sum_{ngeq 1}frac{1}{n^{2}}right)^{2} + sum_{ngeq 1}frac{1}{n^{4}} = frac{pi^{4}}{36} + frac{pi^{4}}{90}
$$
answered Nov 15 at 16:28
Seewoo Lee
5,756825
$$
2S = sum_{ileq j} frac{1}{i^{2}j^{2}} + sum_{igeq j} frac{1}{i^{2}j^{2}} = left(sum_{ngeq 1}frac{1}{n^{2}}right)^{2} + sum_{ngeq 1}frac{1}{n^{4}} = frac{pi^{4}}{36} + frac{pi^{4}}{90}
$$
answered Nov 15 at 16:28
Seewoo Lee
5,756825
answered Nov 15 at 16:28
Seewoo Lee
5,756825
answered Nov 15 at 16:28
Seewoo Lee
5,756825
answered Nov 15 at 16:28
Seewoo Lee
5,756825
5,756825
1
@Seewood Lee Good answer, thanks.
– Mohammad Riazi-Kermani
Nov 15 at 16:32
1
We can generalize the result to $$sum_{ile j}frac{1}{i^r j^r}=frac12left(zeta(r)^2+zeta(2r)right).$$
– Tianlalu
Nov 15 at 16:43
add a comment |
1
@Seewood Lee Good answer, thanks.
– Mohammad Riazi-Kermani
Nov 15 at 16:32
1
We can generalize the result to $$sum_{ile j}frac{1}{i^r j^r}=frac12left(zeta(r)^2+zeta(2r)right).$$
– Tianlalu
Nov 15 at 16:43
1
1
@Seewood Lee Good answer, thanks.
– Mohammad Riazi-Kermani
Nov 15 at 16:32
@Seewood Lee Good answer, thanks.
– Mohammad Riazi-Kermani
Nov 15 at 16:32
1
1
We can generalize the result to $$sum_{ile j}frac{1}{i^r j^r}=frac12left(zeta(r)^2+zeta(2r)right).$$
– Tianlalu
Nov 15 at 16:43
We can generalize the result to $$sum_{ile j}frac{1}{i^r j^r}=frac12left(zeta(r)^2+zeta(2r)right).$$
– Tianlalu
Nov 15 at 16:43
add a comment |
up vote
2
down vote
Suppose that $sum_{i=0}^infty a_i$ is an absolutely convergent series. Then (where $i$ and $j$ range over the nonnegative integers)
$$2sum_{i<=j} a_ia_j=\
2sum_{i<j} a_ia_j+2sum_{i=j} a_ia_j=\
sum_{i<j} a_ia_j+sum_{i>j} a_ia_j+color{green}{2sum_{i=j} a_ia_j}=\
left(sum_{i<j} a_ia_j+color{green}{sum_{i=j} a_ia_j}+sum_{i>j} a_ia_jright)+color{green}{sum_{i=j} a_ia_j}=\
color{red}{{sum_{i,j} a_ia_j}}+color{blue}{sum_{i=j} a_ia_j}=\
color{red}{left(sum_{i} a_iright)^2}+color{blue}{sum_{i} (a_i)^2}.$$
The question here is answered by this identity for $displaystyle a_i={1over i^2}$.
answered Nov 15 at 21:22
Steve Kass
10.8k11429
Very beautiful, thank for the answer..
– Mohammad Riazi-Kermani
Nov 15 at 21:33
add a comment |
up vote
2
down vote
Suppose that $sum_{i=0}^infty a_i$ is an absolutely convergent series. Then (where $i$ and $j$ range over the nonnegative integers)
$$2sum_{i<=j} a_ia_j=\
2sum_{i<j} a_ia_j+2sum_{i=j} a_ia_j=\
sum_{i<j} a_ia_j+sum_{i>j} a_ia_j+color{green}{2sum_{i=j} a_ia_j}=\
left(sum_{i<j} a_ia_j+color{green}{sum_{i=j} a_ia_j}+sum_{i>j} a_ia_jright)+color{green}{sum_{i=j} a_ia_j}=\
color{red}{{sum_{i,j} a_ia_j}}+color{blue}{sum_{i=j} a_ia_j}=\
color{red}{left(sum_{i} a_iright)^2}+color{blue}{sum_{i} (a_i)^2}.$$
The question here is answered by this identity for $displaystyle a_i={1over i^2}$.
answered Nov 15 at 21:22
Steve Kass
10.8k11429
Very beautiful, thank for the answer..
– Mohammad Riazi-Kermani
Nov 15 at 21:33
add a comment |
up vote
2
down vote
up vote
2
down vote
Suppose that $sum_{i=0}^infty a_i$ is an absolutely convergent series. Then (where $i$ and $j$ range over the nonnegative integers)
$$2sum_{i<=j} a_ia_j=\
2sum_{i<j} a_ia_j+2sum_{i=j} a_ia_j=\
sum_{i<j} a_ia_j+sum_{i>j} a_ia_j+color{green}{2sum_{i=j} a_ia_j}=\
left(sum_{i<j} a_ia_j+color{green}{sum_{i=j} a_ia_j}+sum_{i>j} a_ia_jright)+color{green}{sum_{i=j} a_ia_j}=\
color{red}{{sum_{i,j} a_ia_j}}+color{blue}{sum_{i=j} a_ia_j}=\
color{red}{left(sum_{i} a_iright)^2}+color{blue}{sum_{i} (a_i)^2}.$$
The question here is answered by this identity for $displaystyle a_i={1over i^2}$.
answered Nov 15 at 21:22
Steve Kass
10.8k11429
Suppose that $sum_{i=0}^infty a_i$ is an absolutely convergent series. Then (where $i$ and $j$ range over the nonnegative integers)
$$2sum_{i<=j} a_ia_j=\
2sum_{i<j} a_ia_j+2sum_{i=j} a_ia_j=\
sum_{i<j} a_ia_j+sum_{i>j} a_ia_j+color{green}{2sum_{i=j} a_ia_j}=\
left(sum_{i<j} a_ia_j+color{green}{sum_{i=j} a_ia_j}+sum_{i>j} a_ia_jright)+color{green}{sum_{i=j} a_ia_j}=\
color{red}{{sum_{i,j} a_ia_j}}+color{blue}{sum_{i=j} a_ia_j}=\
color{red}{left(sum_{i} a_iright)^2}+color{blue}{sum_{i} (a_i)^2}.$$
The question here is answered by this identity for $displaystyle a_i={1over i^2}$.
answered Nov 15 at 21:22
Steve Kass
10.8k11429
answered Nov 15 at 21:22
Steve Kass
10.8k11429
answered Nov 15 at 21:22
Steve Kass
10.8k11429
answered Nov 15 at 21:22
Steve Kass
10.8k11429
10.8k11429
Very beautiful, thank for the answer..
– Mohammad Riazi-Kermani
Nov 15 at 21:33
add a comment |
Very beautiful, thank for the answer..
– Mohammad Riazi-Kermani
Nov 15 at 21:33
Very beautiful, thank for the answer..
– Mohammad Riazi-Kermani
Nov 15 at 21:33
Very beautiful, thank for the answer..
– Mohammad Riazi-Kermani
Nov 15 at 21:33
add a comment |
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The sum is equal to $sum_nfrac{lceil d(n)/2rceil}{n^2}$, if this helps.
– ajotatxe
Nov 15 at 16:19
According to Mathematica, $sum _{n=1}^{infty } frac{sum _{i=1}^n frac{1}{i^2}}{n^2}={7pi^4over360}$. Maybe this gives you an idea of how to derive the sum.
– Steve Kass
Nov 15 at 16:26
@SteveKass It should be related to $zeta(4)$
– ajotatxe
Nov 15 at 16:28
@SteveKass Thanks for the comment, yes sometimes the answer helps to figure out a solution.
– Mohammad Riazi-Kermani
Nov 15 at 16:28