Closed form for $\sum_{n=2}^{\infty} \big{(}H_{n,n}-1\big{)} $? It is known that $$\sum_{n=2}^{\infty}\big{(}\zeta(n)-1\big{)}=1 .$$
Many more series like these, called rational zeta series, can be evaluated in closed form.
I wonder if we can also obtain similar results for series involving rational sums of generalized harmonic numbers. Such numbers are defined as:
$$H_{n,m} := \sum_{k=1}^{n} \frac{1}{k^{m}} .$$
So they form a finite analogy of zeta values, because $\lim_{n \to \infty} H_{n,m} = \zeta(m). $
Question: can a closed form of the series $$\sum_{n=2}^{\infty} \big{(}H_{n,n}-1\big{)} \approx 0.561 $$ be obtained?
And are any results on "rational generalized harmonic series" known?
 A: We have $$\sum_{n=2}^\infty \sum_{k=2}^n\frac{1}{k^n} = \sum_{k=2}^\infty\sum_{n=k}^\infty\frac{1}{k^n}=\sum_{k=2}^\infty \frac{1}{k^k(1-1/k)}=\sum_{k=2}^\infty \frac{1}{k^{k-1}(k-1)} = \sum_{k=1}^\infty\frac{1}{k(k+1)^k}$$
This is not a closed form for the sum (which I doubt exists) but at least is not a nested sum.
A: Just for the fun
After @jjagmath's answer
$$ \sum_{k=1}^\infty\frac{1}{k(k+1)^k}=0.56119109683529823740260009552415845069306676113\cdots$$ is extremely close to the second positive root of the quintic
$$983 x^5+424 x^4+736 x^3+158 x^2-1015 x+293=0$$ (absolute error of $9.08\times 10^{-21}$)
A: Using the result of @jjagmath we can derive an integral representation of the sum
$$s := \sum_{n=1}^{\infty}(H_{n,n}-1) = \sum_{k=1}^\infty\frac{1}{k(k+1)^k}=\int_{0}^{1} \frac{s^{-s}-1}{\log(\frac{1}{s})}\;ds\tag{1}$$
Indeed, we can write
$$\frac{1}{(1+k)^k}=\frac{1}{\Gamma(k)}\int_{0}^{\infty}t^{k-1} e^{-(k+1)t} \;dt =\frac{1}{\Gamma(k)}\int_{0}^{1} s^k \log(\frac{1}{s})^{k-1}\;ds \tag{2}$$
Inserting this into the second sum of $(1)$ and interchanging integration and summation we have to calculate under the integral a sum which can be evaluated as follows
$$\begin{align}\sum_{k=1}^{\infty} \frac{1}{k} \frac{1}{\Gamma(k)}s^k \log(\frac{1}{s})^{k-1}=\sum_{k=1}^{\infty} \frac{1}{k!} s^k \log(\frac{1}{s})^{k-1}=\frac{1}{\log(\frac{1}{s})}\sum_{k=1}^{\infty} \frac{1}{k!} \left(s \log(\frac{1}{s})\right)^{k}\\
=\frac{1}{\log(\frac{1}{s})}\left(\sum_{k=0}^{\infty} \frac{1}{k!} \left(s \log(\frac{1}{s})\right)^{k}-1\right)=\frac{1}{\log(\frac{1}{s})}\left(e^{s \log(\frac{1}{s})}-1\right)\\
=\frac{1}{\log(\frac{1}{s})}\left(s^{-s} -1\right)\end{align}\tag{3}$$
which is the integrand of $(1)$. This completes the proof of the integral representation in $(1)$.
