I have infinite series:

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

I found the second form by expanding the logarithms in the first series and then combining the terms of the same order.

But I want to obtain a series without any logarithms. Is it possible?

For example, all of the terms have the form $\frac{1}{p 2^m}$, where $m$ is some natural number and $p$ - an odd integer. Can we combine these terms and get some other function, not logarithm?

  • $\begingroup$ if $|x| < 1$ : $\ln(1+x) = \sum_{m=1}^\infty \frac{(-x)^m}{m}$, so $\sum_{k=1}^\infty \frac1k \ln(1+1/2^k) = \sum_{k=1}^\infty \frac1k\sum_{m=1}^\infty \frac{(-1)^m}{m 2^{mk}}$, and you can invert the two sums by the absolute convergent series theorem to get $\sum_{n=1}^\infty \frac{a_n}{2^n}$ with $a_n = \frac{1}{n}\sum_{d | n} (-1)^d$ $\endgroup$ – reuns Feb 23 '16 at 1:27
  • $\begingroup$ Thank you. What does $d|n$ mean? $\endgroup$ – Yuriy S Feb 23 '16 at 1:31
  • 1
    $\begingroup$ I still would be grateful to know how to calculate $a_n$. This is what I was originally asking $\endgroup$ – Yuriy S Feb 23 '16 at 15:34

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