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I was randomly playing around on Desmos, and I realized that this series seems to evidently converge to $\frac{1}{2}$. Is there any reason why this converges to such a nice number?

Please note that I am in high school, so if this is requires an incredibly complex solution I may not understand it (but if it would be of value to others, go ahead)

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    $\begingroup$ If you want to have some fun, you can write $$\begin{align}\sum_{n=1}^\infty (-1)^{n+1}\frac{\operatorname{Im}(e^{in})}{n} &= \operatorname{Im}\sum_{n=1}^\infty (-1)^{n+1}\frac{e^{in}}{n} = -\operatorname{Im}\sum_{n=1}^\infty \frac{(-e^{i})^n}{n} \\&= -\operatorname{Im}\sum_{n=1}^\infty \frac{(-e^{i})^n}{n} = \operatorname{Im} \ln(1+e^{i}) \end{align}$$ and then try to see what can be made rigorous about that. $\endgroup$ – Clement C. Apr 19 '18 at 0:12
  • $\begingroup$ Or you can compute the Fourier series of the function given here and evaluate both sides at $x = \pi$ (I'm not sure if this counts as high school where you are). $\endgroup$ – Mattos Apr 19 '18 at 0:16
  • $\begingroup$ Clement's method is correct. And $\text{Im}\ln(1+e^i) = \frac{1}{2}$ exactly. $\endgroup$ – GEdgar Apr 19 '18 at 0:19
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Because this is the Fourier series for a simple function.

Given any sufficiently nicely-behaved function with period $L$ (this means that for all $x$, $f(x+L)=f(x)$), it is possible to write it as an infinite sum of other periodic functions: in this case, $\sin{(2\pi nx/L)}$ and $\cos{(2\pi nx/L)}$, which also have $L$ as a period. One in fact can find $a_n$ and $b_n$ so that $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos{\left( \frac{2\pi n x}{L} \right)} + b_n \sin{\left( \frac{2\pi n x}{L} \right)}, $$ at any point where the function is continuous (provided $f$ is nice enough). (Yes, this is extremely surprising. When Fourier first proposed it, people found it literally incredible. Mathematics in the nineteenth century took a very long time to sort out satisfactorily how and when Fourier series work.) In particular, the coefficients are given by the integrals $$ a_n = \frac{2}{L} \int_{-L/2}^{L/2} f(x)\cos{\left( \frac{2\pi n x}{L} \right)} \, dx, \qquad b_n = \frac{2}{L} \int_{-L/2}^{L/2} f(x)\sin{\left( \frac{2\pi n x}{L} \right)} \, dx. $$ (You can see that this is at least plausible by multiplying the sum by $\cos{\left( \frac{2\pi m x}{L} \right)}$ or $\sin{\left( \frac{2\pi m x}{L} \right)}$ and integrating over a period: almost all of the trigonometric integrals are easily shown to be zero, leaving only the one with $n=m$ and the same cosine or sine, which then evaluates to $1$.)

In particular, I ask you to believe that the function that is given by extending $f(x) = x$ for $-\pi<x \leq \pi$ to be periodic (so define $f(x+2\pi k)=x$ for every integer $k$) is sufficiently nice that this works. This is called a sawtooth wave. Since this function is odd and cosine is even, all the $a_n$ are zero. On the other hand, $$ \frac{1}{\pi}\int_{-\pi}^{\pi} x\sin{nx} \, dx = \frac{2(-1)^{n+1}}{n}. $$ Therefore, if everything given above is to be believed, for $-\pi<x<\pi$, where the function is continuous and equal to $x$, we have $$ x = \sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n}\sin{nx}. $$ If we now put $x=1$ and divide both sides by $2$, we get exactly your hypothesised value.

Fourier series can be used to evaluate many sums (you can find numerous examples on this site, indeed), and ideas akin to Fourier series are absurdly useful in many branches of modern mathematics (even some ones that don't seem to involve real numbers, like number theory or combinatorics).

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  • $\begingroup$ This is a really well-written answer. +$1$. $\endgroup$ – Clayton Apr 19 '18 at 1:39
  • $\begingroup$ Technically, although the function is simple, it's not a simple function ;-) $\endgroup$ – Daniel Fischer Apr 19 '18 at 11:48
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There are two issues here: We have to prove that the series converges, and then that the sum is ${1\over2}$.

We need the imaginary part of the series $$\sum_{n=1}^\infty(-1)^{n-1}{e^{in}\over n}\ .\tag{1}$$ The general term of $(1)$ is the product of ${1\over n}$, which converges monotonically to $0$, and $(-e^i)^n$. The partial sums of the auxiliary series formed with the second factor, $$\sum_{n=0}^{N-1} (-e^i)^n={1-(-1)^N e^{iN}\over 1+e^i}$$ are bounded. Abel's convergence criterion (a generalization of the theorem on alternating series) then tells us that the series $(1)$ is convergent.

For the sum of $(1)$ we need another theorem of Abel that allows us to make the link to "known facts". The form of the series $(1)$ suggests considering the function $$f(t):=\log\bigl(1+e^i t\bigr)\qquad(-1<t<1)\ .\tag{2}$$ The well known series for $\log(1+x)$ enables us to write $f$ in the form $$f(t)=\sum_{n=1}^\infty(-1)^{n-1}{e^{in}\,t^n\over n}\qquad(-1<t<1)\ .\tag{3}$$ Inspecting $(2)$ we see that $f$ is (complex valued and) continuous in a full neighborhood of $t=1$. On the other hand we have proven above that the series $(3)$ converges also at $t=1$. Abel's second theorem then guarantees that the series $(3)$ with $t:=1$ in fact has the value $f(1)$. This means that $$\sum_{n=1}^\infty(-1)^{n-1}{e^{in}\over n}=f(1)={\rm Log}(1+e^i)=\log\bigl|1+e^i\bigr|+i\,{\rm Arg}(1+e^i)\ .$$ A quick drawing shows that the polar angle of $1+e^i$ is half the polar angle of $e^i=e^{i\,1}$, hence ${1\over2}$.

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