How to prove that the sum of reciprocals of one plus perfect powers is $\frac{\pi^{2}}{3}-\frac{5}{2}$ Let $S$ be Set of perfect powers without duplicates $1,4,8,9,\dots$ (http://oeis.org/A001597 ) How to prove the following?
$$\sum_{s\in S}\frac{1}{s+1}=\frac{\pi^{2}}{3}-\frac{5}{2}$$ (starting with $s=4$ ) I found this formula in the book "Mathematical Constants" by Steven R. Finch on page 113.
 A: Let $S$ be the set of perfect powers without $1$ . That is, $S = \{4,8,9,16,25,27,\ldots\}$. Consider the sum $$
S_{N} = \sum_{s \in S} \frac 1{s^{N}-1}
$$
which converges for $N\geq 2$ because $\sum_{n=1}^\infty \frac 1{n^N}$ converges, and the above sequence is thus a subsequence of a convergent sequence. For $N=1$, a separate argument can be created by "working backwards" from what we do below so I won't really emphasize that point.
We can evaluate this sum using some clever ideas. The first is to consider the set of non-powers $T$ (insist on $1 \notin T$) and its relation to $S$. Of course it is the complement of $S$, but there is a deeper relation.
Indeed, let $s \in S$. We can find $k \geq 2$ such that $s$ is a perfect $k$-th power. Let $K$ be the largest number such that $s$ is a $K$th perfect power. Then, $s^{\frac 1K}$ is a positive integer that has to be a non-power by maximality of $k$.  Thus, every $s \in S$ is uniquely of the form $t^K$ where $K \geq 2$ and $t \in T$. On the other hand, if $t \in T$ and $K \geq 2$, obviously $t^K \in S$. Therefore, we may write $$
\sum_{s \in S} \frac 1{s^N-1} = \sum_{K \geq 2} \sum_{t \in T} \frac{1}{t^{NK}-1}
$$
Now we use a very nice trick : the identity $\frac{1}{n-1} = \frac 1{n} + \frac 1{n(n-1)}$ gives that $$
\sum_{K \geq 2} \sum_{t \in T} \frac{1}{t^{NK}-1} = \sum_{K \geq 2} \sum_{t \in T} \frac 1{t^{NK}} + \sum_{K \geq 2} \sum_{t \in T} \frac 1{t^{NK}(t^{NK} - 1)} \tag{*}
$$
However, observe that $$
\sum_{K \geq 2} \sum_{t \in T} \frac 1{t^{NK}} = \sum_{t \in T} \sum_{ K \geq 2} \frac 1{t^{NK}} = \sum_{t \in T} \frac 1{t^N(t^N-1)}
$$
Therefore, combining this with $(*)$ gives $$
\sum_{K \geq 2} \sum_{t \in T} \frac{1}{t^{NK}-1} = \sum_{t \in T} \frac 1{t^N(t^N-1)} +  \sum_{K \geq 2} \sum_{t \in T} \frac 1{t^{NK}(t^{NK} - 1)}
$$
However, take a very careful look at the RHS here. We are actually summing the quantity $\frac 1{v^N(v^N-1)}$, first for $v \in T$, and then for numbers of the form $v^K$ for $v \in T, k \geq 2$ : which we know to be equal to $S$!
That is, we in fact, have $$
\sum_{t \in T} \frac 1{t^N(t^N-1)} +  \sum_{K \geq 2} \sum_{t \in T} \frac 1{t^{NK}(t^{NK} - 1)} = \sum_{t \in T} \frac 1{t^N(t^N-1)}  +  \sum_{s \in S}\frac 1{s^N(s^N-1)} = \sum_{k=2}^{\infty} \frac 1{k^N(k^N-1)}
$$
We have obtained the identity $$
S_N = \sum_{k=2}^{\infty} \frac{1}{k^N(k^N-1)}
$$
Let's put $N=1$ first. Then, we get by telescoping, $$
S_1 = \sum_{k=2}^{\infty} \frac{1}{k(k-1)} = \sum_{k=2}^{\infty} \left(\frac 1{k-1} - \frac 1{k} \right)\\ = 1 - \frac 12 + \frac 12 - \frac 13+ \ldots = 1
$$
This is a proof of the first identity in Finch's book. The proof of the second identity follows by the evaluation of $S_2$. We write by the telescoping identity $$
S_2 = \sum_{k=2}^{\infty} \frac{1}{k^2(k^2-1)} = \sum_{k=2}^{\infty} \left(\frac 1{k^2-1} - \frac 1{k^2}\right) = \sum_{k=2}^{\infty} \frac 1{k^2-1} - \sum_{k=2}^{\infty} \frac 1{k^2}
$$
We know that $\sum_{k=2}^{\infty} \frac 1{k^2} = \frac{\pi^2}{6}-1$. What about $\sum_{k=2}^{\infty} \frac 1{k^2-1}$? For that, perform partial fractions and notice yet another telescoping occuring.
$$
\sum_{k=2}^{\infty} \frac 1{k^2-1} = \frac 12\sum_{k=2}^{\infty} \frac 2{k^2-1} = \frac 12\sum_{k=2}^{\infty} \left(\frac{1}{k-1} - \frac 1{k+1}\right) \\ = \frac 12 \left(1 - \frac 13 + \frac 12 - \frac 14 + \frac 13 - \frac 15 + \frac 14 - \frac 16 + \ldots\right) \\ = \frac 12\left(1+\frac 12\right) = \frac 34
$$
That is, we obtain $$
S_2 = \frac{3}{4} + 1 - \frac{\pi^2}{6} = \frac{7}{4} - \frac{\pi^2}{6}
$$
We are finally in a position to finish: (and I need to , because merely typing the word telescoping has made my voice hoarse) $$
\sum_{s \in S} \frac 1{s+1} = \sum_{s \in S} \left(\frac{1}{s-1} - \frac{2}{s^2-1}\right) = S_1 - 2S_2 = \frac{\pi^2}{3} - \frac 72 + 1 = \frac{\pi^2}{3} - \frac 52
$$
as desired.

Note that the evaluation of higher $S_N$ is possible, because $$
S_N = \sum_{k=2}^{\infty} \frac 1{k^N-1} - \zeta(N) + 1
$$
One uses partial fraction decomposition, and the definition of the Digamma function like has been done here, to obtain $$
S_N = 1 - \zeta(N) - \frac 1N \sum_{\omega^N = 1}\omega \psi(2-\omega)
$$
