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I have to find the following limit:

$$\lim_{n \to \infty} \sum_{k=1}^n\frac{1}{k\cdot 2^k}$$

The exercise also provides me with the following function definition:

$$ f_n(x) = \frac{x^n}{1-x} $$

I tried to integrate this function by parts on the interval $[0, \frac12]$ and it led me something like this:

$$ I_n = \int_0^\frac12 \frac{x^n}{1-x} \, dx = \log2 - \frac{1}{2^n} + n\int_0^\frac12 x^{n-1} \log(x-1) \, dx $$.

This seems to look like a part of my sum terms but I don't know how to finish the integration and I cannot complete the limit. Could you help me please?

Victor
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    I disagree that this question should be closed. The question Martin suggested does show the value of the series, but does not provide supplements to Victor's try. – choco_addicted Feb 05 '18 at 13:51
  • @choco_addicted well thank you! That's what I am actually trying to find out: how to solve it my way. because I could just go to WolframAlpha to find the value of this... – Victor Feb 05 '18 at 13:58

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