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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.

Elliot Yu
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Mariko
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    Well, Goldbach "showed" that the parallel sum with a minus sign is $1$, see this. Not sure I've seen this one though. – lulu Sep 14 '22 at 20:40
  • For what it's worth, I checked Finch's book. He cites some unpublished notes that I can't find online for this result. – Elliot Yu Sep 14 '22 at 21:19
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    Remark: Adding 1 to the A001597 sequence gives another "registered" sequence A224866 but no mention is made there of the summation of the reciprocal terms of the sequence. – Jean Marie Sep 14 '22 at 21:35
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    Why should this question be closed ? – Jean Marie Sep 14 '22 at 21:36
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    @JeanMarie Some might think it's a little context-thin. I suppose OP could add in the fact that this sum is related to but distinct from the Goldbach-Euler theorem. Though I personally don't think it should be closed, because at least there is a source, and the source itself is pretty bare-bones. – Elliot Yu Sep 14 '22 at 21:45
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    @ElliotYu I don't think it should be closed either. Obviously it's a common ambition for questions that don't try solving themselves, but this is a much harder problem than the norm for that. – J.G. Sep 14 '22 at 21:49
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    @JeanMarie Also I don't think A224866 is A001597 plus one. For example, the former contains $73$ and $109$, but $72$ and $108$ are not perfect powers and not listed in the latter. – Elliot Yu Sep 14 '22 at 22:00
  • To supply one idea, the proof of the Goldbach sum involves rewriting the sum as a double sum over all powers of all non-powers, i.e. $\sum_{s\in S} = \sum_{k\geq 2} \sum_{t\in T}$ where $T = \mathbb{N}^+ - S$. The summand can be expanded into a power series $(t^k - 1)^{-1} = \sum_{i\geq 1} t^{-i k}$, which in the case of a plus sign in the denominator becomes $\sum_{i\geq 1}(-1)^i t^{-i k}$. Then the proof of G-E combines the sum over $i$ with the sum over $t$ into a sum over all integers $n$ greater than $1$, and sum the geometric series over $k$ into $1/(n(n-1))$. – Elliot Yu Sep 14 '22 at 22:31
  • Now in the G-E case, the final sum over $n$ telescopes and gives us $1$. Alternatively, the sum over $n$ is taken first, so we get and identity $\sum_{k\geq 2} (\zeta(k) -1) = 1$. In the flipped sign case, we get an extra sign depending on $n$. Write an integer $n$ as $n = t^i$, where $t$ is a non-power, then the sign is $(-1)^i$. Let's define a function $p$ by $n\mapsto i$ as above. The above procedure then transforms the sum into $\sum_{n\geq 2} (-1)^{p(n)}(n(n-1))^{-1}$. The zeta function expression is not available, because we can no longer sum over $n$ first. – Elliot Yu Sep 14 '22 at 22:32
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    Checking I understand this question, this is meant to be $$\sum_{k=2}^\infty {1\over 2^k+1}+{1\over 3^k+1}+{1\over 5^k+1}+{1\over 6^k+1}+{1\over 7^k+1}+{1\over 10^k+1}+...?$$ – Suzu Hirose Sep 14 '22 at 22:57
  • @SuzuHirose For $2 \leq k < \infty$, but otherwise yes. – Jakob Streipel Sep 14 '22 at 23:01
  • @prets I was just able to edit it! – Suzu Hirose Sep 14 '22 at 23:02
  • @Elliot Yu You are right. – Jean Marie Sep 15 '22 at 06:01

1 Answers1

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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 $$ S_N = 1 - \zeta(N) - \frac 1N \sum_{\omega^N = 1}\omega \psi(2-\omega) $$

Sarvesh Ravichandran Iyer
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