Verify the correctness of $\sum_{n=1}^{\infty}\left(\frac{1}{x^n}-\frac{1}{1+x^n}+\frac{1}{2+x^n}-\frac{1}{3+x^n}+\cdots\right)=\frac{\gamma}{x-1}$

Question

$x\ge2$

$\gamma=0.57725166...$

(1)

$$\sum_{n=1}^{\infty}\left(\frac{1}{x^n}-\frac{1}{1+x^n}+\frac{1}{2+x^n}-\frac{1}{3+x^n}+\cdots\right)=\frac{\gamma}{x-1}$$

Series (1) converges very slowly, we are not sure that (1) has closed form $\frac{\gamma}{x-1}$

Can anyone verify (1)?

Answer 1

For $x =2$ the identity is true.

Claim. $\displaystyle \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{2^n+k} = \gamma. $

Proof. We first rearrange the given sum:

$$ S := \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{2^n+k} = \sum_{l = 2}^{\infty} \bigg( \sum_{\substack{(n,k) \ : \ 2^n + k = l \\ n \geq 1, k \geq 0}} \frac{(-1)^k}{2^n+k} \bigg) = \sum_{l = 2}^{\infty} \frac{(-1)^l}{l} \lfloor \log_2 l \rfloor. \tag{1} $$

(For a careful reader: see below for a rigorous proof.) Then group each $2^n$ terms to write

$$ S = \sum_{n=1}^{\infty} n \bigg( \sum_{l=2^n}^{2^{n+1}-1} \frac{(-1)^l}{l} \bigg) = \sum_{n=1}^{\infty} n ( A_{n-1} - A_n ), $$

where $A_n = H(2^{n+1}-1) - H(2^n-1)$ and $H(n) = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$ is the $n$-th harmonic number. Now the $N$-th partial sum of the RHS is

\begin{align*} \sum_{n=1}^{N} n(A_{n-1} - A_n) &= (A_0 + \cdots + A_{N-1}) - N A_N \\ &= (N-1)H(2^N-1) - N H(2^{N+1}-1). \end{align*}

Taking limit as $N \to \infty$ gives the desired limit $\gamma$. ////

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Remark. Let $f(z) = \sum_{k=0}^{\infty} \frac{(-1)^k}{z+k}$.

1. Using the formula $f(z) = \psi_0(z) - \psi_0(z/2) - \log 2$ together with the asymptotic expansion of $\psi_0(z)$ gives a much shorter proof.

2. We can check that $f(z) = \frac{1}{2z} + \mathcal{O}(z^{-2})$. Thus $$ \sum_{n=1}^{\infty} f(x^n) = \frac{1}{2(x-1)} + \mathcal{O}\left(\frac{1}{x^2} \right). $$ This tells us that the proposed formula is false, which is already confirmed by other users.

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Addendum 1 - A careful justification of $\text{(1)}$. The trick is to group each pair of successive terms to create a non-negative sum. Then by the Tonelli's theorem we can freely rearrange the order of summation.

\begin{align*} \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{2^n+k} &= \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{1}{(2^n+2k)(2^n+2k+1)} \\ &= \sum_{l' = 1}^{\infty} \bigg( \sum_{\substack{(m,k) \ : \ 2^m + k = l' \\ m \geq 0, k \geq 0}} \frac{1}{(2^{m+1}+2k)(2^{m+1}+2k+1)} \bigg) \\ &= \sum_{l' = 1}^{\infty} \frac{1 + \lfloor \log_2 l' \rfloor}{2l'(2l'+1)} \\ &= \sum_{l' = 1}^{\infty} \bigg( \frac{\lfloor \log_2 (2l') \rfloor}{2l'} - \frac{\lfloor \log_2 (2l'+1) \rfloor}{2l'+1} \bigg)\\ &= \sum_{l = 1}^{\infty} \frac{(-1)^l}{l}\lfloor \log_2 l \rfloor. \end{align*}

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Addendum 2 - Heuristic. We use the following asymptotic expansion

$$ \sum_{k=0}^{\infty} \frac{(-1)^k}{z+k} = \psi_0(z) - \psi_0(z/2) - \log 2 = \frac{1}{2z} + \sum_{k=1}^{\infty} \frac{B_{2k}}{2k} \frac{2^{2k}-1}{z^{2k}}, $$

where $(B_k)$ are Bernoulli numbers. (Notice: This is only a formal sum because the asymptotic expansion above does not converge for any $z$. It is numerically meaningful only when we truncate the sum.) Then we formally have

\begin{align*} \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{x^n + k} &= \frac{1}{2(x-1)} + \sum_{n=1}^{\infty} \frac{B_{2n}}{2n} \frac{2^{2n}-1}{x^{2n}-1} \\ &= \frac{1}{2(x-1)} + 2 \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \Gamma(2n)\zeta(2n)}{(2\pi)^{2n}} \frac{2^{2n}-1}{x^{2n}-1} \\ &= \frac{1}{2(x-1)} + 2 \int_{0}^{\infty} \bigg( \sum_{n=1}^{\infty} (-1)^{n-1}\zeta(2n) \frac{2^{2n}-1}{x^{2n}-1} t^{2n-1} \bigg) e^{-2\pi t} \, \mathrm{d}t. \end{align*}

Although the summation in the last line does not converge for $t \geq x/2$, it may have an analytic continuation on all of $t > 0$. Then with that continuation the last expression may make sense and even possibly give the correct formula for the original sum. Indeed, when $x = 2$ it follows that

$$ 2 \sum_{n=1}^{\infty} (-1)^{n-1}\zeta(2n) t^{2n-1} = \pi \coth \pi t - \frac{1}{t} $$

and hence

$$ \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{2^n + k} \text{ “}=\text{" } \frac{1}{2} + \int_{0}^{\infty} \bigg( \pi \coth \pi t - \frac{1}{t} \bigg) e^{-2\pi t} \, \mathrm{d}t = \gamma. $$

Now we know that this alleged equality is indeed true. So the above formal computation seems to give the correct answer (at least for $x =2$, and hopefully for $x > 2$). But for $x > 2$, I have no idea what will come out.

Answer 2

You wish to try and see if the following is true:

$$\sum_{n=1}^\infty\sum_{k=0}^\infty\frac{(-1)^k}{k+x^n}=\frac\gamma{x-1}$$

For $x=0$,

$$\sum_{n=1}^\infty\sum_{k=0}^\infty\frac{(-1)^k}{k}=-\gamma$$

$$\sum_{k=0}^\infty\frac{(-1)^k}{k}=undefined\tag{$k=0$}$$

So... no, they are not equal. I do not think that the LHS should equal the right only for $x\ge2$ because of analytical continuation.

WolframAlpha says the following:

$$\sum_{n=1}^\infty\sum_{k=0}^\infty\frac{(-1)^k}{k+x^n}=\sum_{k=0}^\infty\sum_{n=1}^\infty\frac{(-1)^k}{k+x^n}$$

$$=\sum_{k=0}^\infty\frac{2\psi_x^{(0)}\left(-\frac{\ln(-\frac kx)}{\ln(x)}\right)+2\ln(-\frac kx)+2\ln(x-1)+\ln(x)}{2k\ln(x)}$$

Unfortunately, wolfram alpha cannot take the sum of that. However, I note here, that for any value of $x$, this is undefined, since we must first evaluate it for $k=0$. It is undefined for the numerator and denominator.

So... let's ignore that. Let's assume we can ignore $k=0$ for later.