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Using the Taylor-Maclaurin series and differentiation/integration calculate the infinite sum n/((n+1)(2^n)) from n=1 to infinity. I have tried to write it as ∑ 1/2ⁿ - ∑ 1/((n+1)*2ⁿ) but still cannot solve the second sum and get the final answer which is supposed to be 2-2ln2 (according to some calculators). Any help would be greatly appreciated.

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  • $\begingroup$ The second sum is found using the Maclaurin series of $\ln{(1-x)}$ evaluated at $x=1/2$. $\endgroup$ Jun 13 '20 at 15:56
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hint

Consider the power series $$\sum x^n$$ for $x\in (-1,1)$, we have $$\sum_{n=0}^{+\infty}x^n=\frac{1}{1-x}$$

by integration, for $ x\in (-1,1) $, $$\sum_{n=0}^{+\infty}\frac{x^{n+1}}{n+1}=\int_0^x\frac{dt}{1-t}=-\ln(1-x)$$

If $ x=\frac 12$, it gives

$$\frac 12\sum_{n=0}^{+\infty}\frac{1}{2^n(n+1)}=\ln(2)$$

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  • $\begingroup$ Thank you very much $\endgroup$
    – Bora Kutlu
    Jun 13 '20 at 16:26

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