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This came up while integrating: \begin{align}\int_0^1\frac{dx}{7^{[\frac{1}{x}]}}\end{align} after transformation $\frac{1}{x}=u$, it became \begin{align}\int_1^{\infty}\frac{du}{u^27^{[u]}}=\sum_1^{\infty}\frac{1}{7^n}(\frac{1}{n}-\frac{1}{n+1})\end{align} Proceeding after this is unclear. Due to the source of the question, there is a hint available, but it would be appreciated if the answerer does not use the hint so as to see how we evaluate the summation without any further information

Here is a hint

the closed form includes logarithmic terms

Shimura Variety
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    Hint: Find the sum of $\sum^\infty_{n=1}x^n/n$...something to do with integrating geometric series. – Jacky Chong Aug 09 '22 at 05:49
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    And then indeed, $\sum_{n=1}^\infty \frac1{7^n}\frac1{n+1} = \sum_{m=2}^\infty \frac1{7^{m-1}}\frac1m = 7 \sum_{m=1}^\infty \frac1{7^m} \frac1m - 1$, so the same evaluation will help you with both halves. – Greg Martin Aug 09 '22 at 05:53

1 Answers1

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$-\log(1-x)=\displaystyle \sum_{k=1}^{\infty} \frac{x^k}{k}, |x|<1$ Integrate.w.r.t. in $(0,t)$ to get $$t+(1-t)\log(1-t)=\sum_{k=0}^{\infty} \frac{t^{k+1}}{k(k+1)}$$ Put $t=1/7$ to get $$\sum_{k=1}^{\infty} \frac{1}{7^k k(k+1)}=1-6 \log(7/6)$$

MathDona
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