- #36

Fred Wright

- 365

- 215

I don't follow what you don't follow. $$k=j=nI cannot follow you. In your very first equation I get different terms:

$$

\sum_{k,j=^1}^\infty \dfrac{1}{kj(k+j)^2}=\underbrace{\dfrac{1}{1\cdot 2^2}}_{(1,1)}+\underbrace{\dfrac{2}{2\cdot 3^2}}_{(1,2)}+\underbrace{\dfrac{1}{4\cdot 4^2}}_{(2,2)}+\underbrace{\dfrac{2}{3\cdot 4^2}}_{(1,3)}+\underbrace{\dfrac{2}{6\cdot 5^2}}_{(2,3)}+\underbrace{\dfrac{1}{9\cdot 6^2}}_{(3,3)}+\ldots

$$

so you must have used a summation I cannot see. And the result is a different one.

The trick is indeed to rewrite the summands. My proof uses integrals.

\\ \sum_{k,j=1}^{\infty}\frac{1}{kj(k+j)^2}=\sum_{n=1}^{\infty}\frac{1}{n^2(2n)^2}=\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^4}$$In my post on this problem I claim,$$\sum_{k,j=1}^{\infty}\frac{1}{kj(k+j)^2}=\frac{\pi^4}{360}$$I ran the following C code which iterates the sum 10000 times:

double z, zsum = 0, dz;

for (int i = 1; i < 10000; i++)

{

dz = (double)i;

z = 1.0 / pow(dz, 4);

zsum += z;

}

zsum *= 0.25;

The zsum variable agrees with ##\frac{\pi^4}{360}## computed on my HP calculator to 10 decimal places. QED I think. I request "solved by" honor.