How to evaluate $\sum\limits_{n=1}^{+\infty}\frac{1}{1^k+2^k+\cdots+n^k}$?

There is post here asking for $$\sum_{n=1}^{+\infty}\frac{1}{1^2+2^2+\cdots+n^2}.$$ And the answer is $18-24\ln 2$.

It is easy to elvaluate that $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{1+2+\cdots+n}=2$ and it is not difficult to justify the convergence of $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{1^k+2^k+\cdots+n^k}$ for all $k>1$.

Here comes my question: How to evaluate $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{1^3+2^3+\cdots+n^3}$ or in general, what is $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{1^k+2^k+\cdots+n^k}$ if $k$ is a positive integer.

I guess the way is pretty much the same, as there are always formulas to evaluate $$\sum_{i=1}^n i^k, $$ For $k=3$ the result of the sum above is $$\frac{4}{3} (-9+\pi^2)$$ For higher terms you will probably not always get a nice result, but for $k=5$ you have $$ 4\left(15-\pi^2 + 2\sqrt{3} \pi \tan\left(\frac{\sqrt{3}\pi}{2}\right)\right)$$

I will calculate the $k=3$ case.
As $$\sum_{i=1}^n i^3 = \frac{1}{4} n^2 (1+n)^2$$ we have \begin{align*} \sum_{n=1}^\infty \frac{1}{\sum_{i=1}^n n^3 } &= 4 \sum_{n=1}^\infty \frac{1}{m^2(1+m)^2}\\ &=\sum_{n=1}^\infty \frac{1}{n^2} - \frac{2}{n} + \frac{2}{n+1} + \frac{1}{(n+1)^2}\\ &=4\sum_{n=1}^\infty \frac{1}{n^2} +4\sum_{n=1}^\infty \frac{1}{(1+n)^2} - 8 \cdot \sum_{n=1}^\infty \frac{1}{n} -\frac{1}{n+1}\\ &=4\left(\frac{\pi^2}{6} +\frac{\pi^2}{6}-1-2\right)\\ &=\frac{4}{3}(\pi^2-9)\end{align*}

Though it may be tricky to find a general formula, we get with Faulhaber's formula that

$$\sum_{n=1}^\infty \frac{1}{\sum_{m=1}^n m^k}= \sum_{n=1}^\infty \frac{1}{\sum_{j=0}^k\binom{k}{j}\frac{B_{k-j}}{j+1}n^{j+1}}=$$

where $B_k$ are Bernoulli numbers.