Partially Accepting

So, the problem is actually -

$\displaystyle \sum_{i=1}^{\infty}{\left(\frac{1}{{i}^{4}}\displaystyle \sum_{k=1}^{i}{\frac{1}{{k}^{4}}}\right)}$

Well, that looks easy now, right? Well, yeah.

We are going to transform the sum such that it becomes easy for us to compute.

$\displaystyle \sum_{n=1}^{\infty}{\frac{\left \lfloor \frac{\tau(n)}{2} \right \rfloor}{{n}^{4}}}$

where $\tau(n)$ is the divisor function .

You may see how that might have come. It came out, however, because we are multiplying two numbers whatsoever to get a new number and then summing all. So, the number of divisors a number has and the number of times it will come out in the sum would be just half. Also, that is not the case with squares where number of divisors is odd.

So, after we have got why and how that is so, it is easy to remove that floor(since we(more specifically I) don't know how to compute directly from the floor function).

$\displaystyle \frac{1}{2} \sum_{n=1}^{\infty}{\frac{\tau(n)}{{n}^{4}}} + \frac{1}{2} \sum_{n=1}^{\infty}{\frac{1}{{({n}^{2})}^{4}}}$

The first sum can be written as $\displaystyle \sum_{n=1}^{\infty}{\frac{1}{{n}^{4}} \left(\sum_{p=1}^{\infty}{\frac{1}{{p}^{4}}} \right)} = {\zeta(4)}^{2}$

And the second of course becomes $\zeta(8)$ .

So, our total sum becomes - $\displaystyle \frac{1}{2} \left ({\zeta(4)}^{2} + \zeta(8)\right)$

$\displaystyle = \frac{13 {\pi}^{8}}{113400}$