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I'm currently tackling a homework problem that asks if the following series converges or diverges: $$\sum_{n=0}^\infty \frac{i^n}{n+1} $$ Currently, I have attempted the ratio test to no avail, because I reach a final expression of $\lim\limits_{n \to \infty} \frac{i(n+1)}{n+2}$ which doesn't really help in determining convergence. Also, I have tried to think of similar series that could potentially be used to do a comparison test, but I can't think of anything that works. Any answers and explanations would be very useful and very appreciated!

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  • $\begingroup$ Hint: look at the real and imaginary parts as separate series. $\endgroup$ – dxiv Feb 8 '17 at 5:44
  • $\begingroup$ @dxiv can you clarify what exactly is the "real part" of this series? To me, the element of the series is strictly imaginary. $\endgroup$ – linearalgebrathrowaway Feb 8 '17 at 5:47
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    $\begingroup$ Write down the first few terms: $$\frac{1}{1} + \frac{i}{2}-\frac{1}{3}-\frac{i}{4}+ \frac{1}{5} + \cdots$$ $\endgroup$ – dxiv Feb 8 '17 at 5:49
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$$\sum_{n=0}^\infty \frac{i^n}{n+1}=\frac{1}{0+1}+\frac{i}{1+1}+\frac{-1}{2+1}+\frac{-i}{3+1}+\cdots$$ $$=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}+i\sum_{n=0}^\infty \frac{(-1)^n}{2n+2}$$
Do you recognize the first sum from anywhere? Maybe the Leibnez Formula for $\pi$? What about the second sum? If we take a factor of $2$ out of the denominator, do you notice a famous series for $\log(2)$?

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\begin{eqnarray} -\log(1-z)&=&\int\frac{1}{1-z}dz\\&=&\int\sum_1z^n\\&=&\sum_1\int z^n\\&=&\sum_1\frac{z^{n+1}}{n+1}\\&=&z\sum_1\frac{z^{n}}{n+1}\\&=&z\sum_0\frac{z^{n}}{n+1}-z \end{eqnarray} set $z=i$, we have $1+i\log(1-i)=\color{blue}{1+\dfrac{\pi}{4}+2k\pi+i\dfrac12\ln2}$.

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