I'm trying to calculate the following series: $$ \sum_{n=1}^\infty \frac{1}{3^n-2^n}=\frac{1}{3-2}+\frac{1}{9-4}+\ldots $$ or the general case: $$ \sum_{n=1}^\infty \frac{1}{(x+1)^n-x^n} $$ I have tried some methods but still have no idea, is there any way to solve such series?
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What is the source of this question? – Shreyansh Pathak Oct 06 '19 at 14:50
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A friend told me, it's just his inadvertently thought. – 3usi9 Oct 06 '19 at 14:54
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It is approximately 1.27498 – MafPrivate Oct 06 '19 at 14:55
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What are the other methods you used? – Shreyansh Pathak Oct 06 '19 at 14:56
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I tried to expand it with binomial expression, but can't simplify the result. Mathematica told me $$\sum _{n=1}^{\infty } \frac{1}{4^n-2^n}$$ exists and equals to $$\frac{-2 \psi _2^{(0)}(1)-\ln (2)}{\ln (4)}$$. – 3usi9 Oct 06 '19 at 15:00
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Mathematica gives closed forms for $$\sum_{n=1}^{\infty}\frac{1}{a^{mn}-a^{(m-1)n}},$$ where $a\in\mathbb{Z}{\geq2}$ and $m\in\mathbb{Z}{\geq1}$, in terms of logarithms and q-Polygamma functions – MathIsFun7225 Oct 06 '19 at 15:13
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Already asked today Can the result of $\sum\limits_{n=1}^{\infty}\dfrac{1}{3^n-2^n}$ be expressed as a closed form? – Sil Oct 06 '19 at 15:20
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It is $$\sum_{n=1}^\infty \frac{1}{3^n-2^n}\approx 1.27498$$ – Dr. Sonnhard Graubner Oct 06 '19 at 15:33