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Prove that the series $\displaystyle \sum_{n=1}^{\infty}\frac{i^n}{n}$ converges.

Optional. find it's sum, if possible.

Comments. I am aware of the general result about the convergence (not absolute) of $\displaystyle \sum_{n=1}^{\infty}\frac{z^n}{n}$, for $z\neq 1$. But i feel that the answer to the above problem can be more trivial, although have not made any interesting approach so far.

Thank you in advance!

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    $\begingroup$ Look at the even $n$ terms, which would sum up as the real part, and then look at the odd $n$ terms, which would sum up as the imaginary part. Show that both real part and imaginary part converge. $\endgroup$ – edm Sep 23 '17 at 17:36
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We have $\frac{1}{1-z}=\sum_{n\geq 0}z^n$ for any $z\in\mathbb{C}$ such that $|z|<1$. The convergence is uniform over any compact subset of $\{z\in\mathbb{C}:|z|<1\}$, hence we are allowed to state

$$ \int_{0}^{i}\frac{dz}{1-z} = \sum_{n\geq 0}\frac{i^{n+1}}{n+1} = \sum_{n\geq 1}\frac{i^n}{n} $$ where the LHS equals $$ -\log(1-i) = -\log\left(\sqrt{2}\, e^{-\frac{\pi i}{4}}\right) = \color{red}{-\frac{\log 2}{2}+\frac{\pi i}{4}}.$$ The convergence of the original series is granted by Dirichlet's test, as already remarked by other users. You may also notice that the real part is given by the terms of the original series with $n\in 2\mathbb{N}$ and the imaginary part is given by the terms of the original series with $n\in 2\mathbb{N}+1$. On the other hand, $$ \sum_{n\geq 1}\frac{(-1)^n}{n}=-\log(2),\qquad \sum_{n\geq 0}\frac{(-1)^n}{2n+1}=\frac{\pi}{4} $$ are pretty well-known. They can be proved with the same approach, restricted to the real line only.

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  • $\begingroup$ Sorry for my question but where is the implication that this can also be used for $z=i$, which has $|i| = 1$? Is it because of Dirichlet's test? $\endgroup$ – MrYouMath Sep 23 '17 at 17:46
  • $\begingroup$ @MrYouMath: the first equality holds for any $z$ close enough to the origin and it can be termwise-integrated due to uniform convergence. The norm of $i$ is $1$. $\endgroup$ – Jack D'Aurizio Sep 23 '17 at 17:48
  • $\begingroup$ The first equality is clear. I mean the implication that it is also a valid representation for $z=i$ which is not $|z|<1$. $\endgroup$ – MrYouMath Sep 23 '17 at 17:52
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    $\begingroup$ @MrYouMath: it is not, $\sum_{n\geq 1}i^n$ is not convergent. But I do not use an evaluation of $\sum_{n\geq 0}z^n$ at $z=i$, I use an evaluation of the primitive $\sum_{n\geq 0}z^{n+1}/(n+1)$ at $z=i$, and I am allowed to do it by the given reasons. $\endgroup$ – Jack D'Aurizio Sep 23 '17 at 18:01
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HINT: Notice that $$\sum_{n=1}^\infty \frac{i^n}{n}=\sum_{n=1}^\infty \frac{(-1)^n}{2n}-i\sum_{n=1}^\infty \frac{(-1)^n}{2n-1}$$

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  • $\begingroup$ I don't think that rearranging is possible for the infinite sum as it is not absolutely convergent. But in essence, this should lead to the solution. $\endgroup$ – MrYouMath Sep 23 '17 at 17:40
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    $\begingroup$ @MrYouMath The rearrangement is valid here, you merely need to check the partial sums of each side and make sure they can match up. $\endgroup$ – Simply Beautiful Art Sep 23 '17 at 17:59
  • $\begingroup$ @SimplyBeautifulArt: I know it is possible in this case but it would be better to be more precise to explicitly state that. The answer, as it is now, is not good without further explanation. Others might think that you can always rearrange infinite sums as you like. $\endgroup$ – MrYouMath Sep 23 '17 at 18:01
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Hint: Look at ther partial sums is equal to:

$$\frac{i}{1}-\frac{1}{2}-\frac{i}{3}+\frac{1}{4}+\ldots \frac{i^n}{n}$$

Factor the complex part and the imaginary part. Notice that you have alternating series in the real and imaginary part. Both series converge by the Leibniz criterion. Additionally compare the real and imaginary part with the Taylor series of $\arctan(x)$ and $\ln(1+x)$.

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  • $\begingroup$ Excellent comment, regarding the non-absolute convergence and the rearrangement of the given series. $\endgroup$ – Nikolaos Skout Sep 23 '17 at 21:23
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    $\begingroup$ @NikolaosSkout Thank you. Seems like we are the only users worried about rearrangement :D. $\endgroup$ – MrYouMath Sep 23 '17 at 21:28
  • $\begingroup$ You are welcome. It is just that sometimes things work...! $\endgroup$ – Nikolaos Skout Sep 23 '17 at 21:50
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You can use Dirichlet's test. Of course if $|z|<1$ then $$\sum_{n=1}^\infty\frac{z^n}n=\ln\frac1{1-z}$$ (for a suitable branch of the complex logarithm).

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$$\sum_{n=1}\dfrac{i^n}{n}=\sum_{n=1}\dfrac{\cos\frac{n\pi}{2}}{n} + i\sum_{n=1}\dfrac{\sin\frac{n\pi}{2}}{n}$$ both converge with Dirichlet!

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Take a look at the following picture :

enter image description here

and see the connection with the issue in the so-called Argand plane (knowing that what is represented is a partial sum of the series).

It remains to prove that this "inward spiraling" movement terminates as a limit point... therefore, I don't say that this graphics constitue a "proof without words" ; it just illustrates the plausibility of a convergence...

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