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Analyze the convergence of the series $\sum_{n=1}^{\infty} \frac{n}{(2i)^{n}}$.

Applying Dirichlet's test with $a_n=\frac{n}{2^n}$ and $b_n=\frac{1}{i^n}=(-i)^n$, I can show that the series converges. Using mathematica I know that the series converge to $-\frac{8}{25}-\frac{6}{25}i$. How can I find this number without using mathematica?

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Your series is$$\sum_{n=1}^\infty nz^n\text,$$with $z=\frac1{2i}=-\frac i2$. But$$\sum_{n=1}^\infty nz^n=z\sum_{n=1}^\infty nz^{n-1}=z\left(\sum_{n=0}^\infty z^n\right)'=\frac z{(1-z)^2}$$and when $z=-\frac i2$ this is equal to $-\frac8{25}-\frac6{25}i$.

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  • $\begingroup$ and also $|z|=1/2<1$ $\endgroup$ – G Cab Jun 16 '17 at 9:13

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