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Define $\gamma(t)$ as $$\gamma(t) = \sum_{n=1}^\infty \sum_{k=2^n}^\infty \frac{(-1)^k}{k+t}. $$ We have $\gamma(0)=\gamma = 0.577....$ I want to find a closed form when $t>0$ is an integer. I believe $\gamma(1)= 1-\gamma$.

I found this paper by Sondow and Zudilin which is possible to find this series but I'm having difficult following it. https://arxiv.org/pdf/math/0304021.pdf

Another representation for the double sum is: $$ \gamma(t) = \sum_{n=1}^\infty \int_0^1 \frac{x^{2^n}x^{t-1}}{1+x}dx$$

Pinteco
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1 Answers1

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Let's define $\;d(t):=\gamma(t+1)+\gamma(t)\;$ then : \begin{align} d(t)& = \sum_{n=1}^\infty \sum_{k=2^n}^\infty \frac{(-1)^k}{k+t+1}+\sum_{n=1}^\infty \sum_{k=2^n}^\infty \frac{(-1)^k}{k+t}\\ & = \sum_{n=1}^\infty \sum_{k=2^n+1}^\infty \frac{(-1)^{k-1}}{k+t}+\sum_{n=1}^\infty \sum_{k=2^n}^\infty \frac{(-1)^k}{k+t}\\ & = \sum_{n=1}^\infty \frac{(-1)^{2^n}}{2^n+t}\\ \tag{1}& = \sum_{n=1}^\infty \frac 1{2^n+t}\\ \end{align}

You may conclude that indeed $\gamma(1)=1-\gamma\;$ (since $\;d(0)=1$)
and obtain the next terms using $(1)$ or the $\frac 12$-digamma function : $$\tag{2} d(t)= \sum_{n=1}^\infty \frac 1{2^n+t}= \frac{\psi_{1/2}(-\log_2(-t/2))}{t \log(2)}-\frac 1t$$

Raymond Manzoni
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  • Mathematica is giving a complex number. Probably because log(-t). Why this sum is not real valued? – Pinteco Oct 28 '23 at 10:55
  • You may consider $(1)$ too as a Lambert-series (see for example this ). Anyway $(2)$ was provided by Alpha which seems in pain with q-digamma...) – Raymond Manzoni Oct 28 '23 at 10:59
  • (the problem as indicated in the accepted answer is that these expressions are rather for $\dfrac 1{2^n-t}$ than for $\dfrac 1{2^n+t}$ so that imaginary numbers will be used to change the sign. The same way $2$ will become $1/2$ because of the usual $2^n$ at the numerator of Lambert series. Shortly simple expressions for $(1)$ are hard to find in the litterature). See too this. – Raymond Manzoni Oct 28 '23 at 11:19