# Learn Further Sums Found Through Fourier Series

In an earlier insight, I looked at the Fourier series for some simple polynomials and what we could deduce from those series. There is a lot more to be found, however.

## Evaluating the Fourier series at π/2

### First series

In the last insight, I showed that if f(x) = x for -π<x<πand f(-π)=f(π)=0, the general  term of the series was $a_{n}=\frac{(-1)^{n}}{in}$ with $a_{0}=0$.  Since f(x) is differentiable at π/2, the Fourier series converges at that point, giving $$\frac{\pi}{2}=\sum_{n=-\infty}^{-1}\frac{(-1)^{n}}{in}e^{in\frac{\pi}{2}}+0+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{in}e^{in\frac{\pi}{2}}$$

Changing n to –n in the first sum, we get$$\frac{\pi}{2}=\sum_{n=1}^{\infty}\frac{(-1)^{n}}{-in}e^{-in\frac{\pi}{2}}+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{in}e^{in\frac{\pi}{2}}$$

Now $\frac{e^{inx}-e^{-inx}}{2i}=\sin(nx)$ and therefore $\frac{\pi}{2}=2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\sin(\frac{n\pi}{2})$. Since $\sin(\frac{n\pi}{2})$ is 0 for all even multiples of p/2 and alternating between +1 and -1 for odd multiples, we have $$\frac{\pi}{4}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}$$

Not exactly a new result, it is usually known as Leibniz’ formula, but it shows that our calculations are correct.

A bit more unusual: The result can also be written as $\frac{\pi}{2}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n-\frac{1}{2}}$. Compare this result with the well-known fact that $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\ln(2)$…

### Second series

In the last insight, I used f(x) = x2 for -π≤x≤π. The general term was then $a_{n}=\frac{2}{n^{2}}$ with $a_{0}=\frac{\pi^{2}}{3}$. Inserting x=π/2 gives $$\frac{\pi^{2}}{4}=\sum_{n=-\infty}^{-1}\frac{2}{n^{2}}e^{in\frac{\pi}{2}}+\frac{\pi^{2}}{3}+\sum_{n=1}^{\infty}\frac{2}{n^{2}}e^{in\frac{\pi}{2}}$$

Change n to –n in the first sum and reorder: $$\frac{\pi^{2}}{4}-\frac{\pi^{2}}{3}=\sum_{n=1}^{\infty}\frac{2}{n^{2}}e^{-in\frac{\pi}{2}}+\sum_{n=1}^{\infty}\frac{2}{n^{2}}e^{in\frac{\pi}{2}}$$

Now $\frac{e^{inx}+e^{-inx}}{2}=\cos(nx)$ and therefore $$\frac{-\pi^{2}}{12}=4\sum_{n=1}^{\infty}\frac{1}{n^{2}}\cos(\frac{n\pi}{2})$$.

Since $\cos(\frac{n\pi}{2})$ is 0 for all odd multiples of π/2 and alternating between +1 and -1 for even multiples, we have $$\frac{-\pi^{2}}{48}=4\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}}$$

Change sign and reorder: $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{2}}=\frac{\pi^{2}}{12}$$

Again, a known fact, but it verifies our procedure.

Staying with the inverse squares, let us try a slightly more unusual function.

Let f(x) be periodic with period 2π where f(x) = cosh(x) for -π≤x≤π.  Then $$a_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\cosh(t)e^{-int}dt=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1}{2}(e^{t}+e^{-t})e^{-int}dt$$$$=\frac{1}{4\pi}\int_{-\pi}^{\pi}(e^{t-int}+e^{-t-int})dt$$$$=\frac{1}{4\pi}(\frac{1}{1-in}(e^{\pi-in\pi}-e^{-\pi+in\pi})-\frac{1}{1-in}(e^{-\pi-in\pi}-e^{\pi+in\pi}))$$

Remembering that $e^{-in\pi}=e^{in\pi}=(-1)^{n}$,  we have $$a_{n}=\frac{1}{4\pi}(\frac{1+in}{1+n^{2}}(-1)^{n}(e^{\pi}-e^{-\pi})-\frac{1-in}{1+n^{2}}(-1)^{n}(e^{-\pi}-e^{\pi})$$$$=\frac{(-1)^{n}}{2\pi}(\frac{1+in}{1+n^{2}}(-1)^{n}\sinh(\pi)+\frac{1-in}{1+n^{2}}(-1)^{n}\sinh(\pi))$$$$=\frac{(-1)^{n}}{\pi}(\frac{\sinh(\pi)}{1+n^{2}})$$

For n=0 we have $$a_{0}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\cosh(t)dt=\frac{1}{2\pi}(\sinh(\pi)-\sinh(-\pi))=\frac{\sinh(\pi)}{\pi}$$

Thus $$\cosh(x)=\sum_{n=-\infty}^{-1}\frac{(-1)^{n}}{\pi}\frac{\sinh(\pi)}{1+n^{2}}e^{inx}+\frac{\sinh(\pi)}{\pi}+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\pi}\frac{\sinh(\pi)}{1+n^{2}}e^{inx}$$$$=\frac{\sinh(\pi)}{\pi}(1+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{1+n^{2}}(e^{inx}+e^{-inx}))$$

For  this tells us (the convergence of the series is trivial) that$$\cosh(\pi)=\frac{\sinh(\pi)}{\pi}(1+\sum_{n=1}^{\infty}\frac{2}{1+n^{2}}$$

Reordering this expression, we get $$\sum_{n=1}^{\infty}\frac{1}{n^{2}+1}=\frac{1}{2}(\pi\coth(\pi)-1)$$

If we evaluate the series for $x=0$ instead, we get $$\cosh(0)=1=\frac{\sinh(\pi)}{\pi}(1+\sum_{n=1}^{\infty}\frac{2\cdot (-1)^{n}}{1+n^{2}}$$

Reordering this expression, we get $$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{1+n^{2}}=\frac{1}{2}(\frac{\pi}{\sinh(\pi)}-1)$$

Adding 1 to both sides of the expression, we get$$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n^{2}+1}=\frac{1}{2}(\frac{\pi}{\sinh(\pi)}+1)$$

## Third series

In the last insight, I used f(x) = x32x. Here I will just change the sign and use $f(x)=\pi^{2}x-x^{3}$. It will, of course, change the sign of the calculated coefficients, so $a_{n}=\frac{-6}{in^{3}}$ with $a_{0}=0$. Inserting x=π/2 gives $$\frac{\pi^{3}}{2}-\frac{\pi^{3}}{8}=\sum_{n=-\infty}^{-1}\frac{-6}{in^{3}}e^{in\frac{\pi}{2}}+\sum_{n=1}^{\infty}\frac{-6}{in^{3}}e^{in\frac{\pi}{2}}$$

Reordering: $$\frac{\pi^{3}}{16}=\sum_{n=1}^{\infty}\frac{1}{in^{3}}e^{-in\frac{\pi}{2}}+\sum_{n=1}^{\infty}\frac{-1}{in^{3}}e^{in\frac{\pi}{2}}$$

Again, we have a sine function, so $$\frac{\pi^{3}}{32}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^{3}}$$

Reducing the denominator: $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n-\frac{1}{2})^{3}}=\frac{\pi^{3}}{4}$$

These results may be new, I do not know (I certainly have not seen them before).

Read my next article: Why Road Capacity Is Almost Independent of the Speed Limit

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5 replies
1. Svein says:
The Electrician

Did you notice that Mathematica disagrees on the sign of the last one?No, saw it just now. Mathematica is correct, the correct exponent for (-1) should be (n-1), not n.

2. The Electrician says:
Svein

Then Mathematica and I agree. Fine!

For me this is part of the road I am currently going. As long as I am getting correct answers along the way, I am happy.Did you notice that Mathematica disagrees on the sign of the last one?

3. Svein says:
The Electrician

Mathematica gave this:Then Mathematica and I agree. Fine!

For me this is part of the road I am currently going. As long as I am getting correct answers along the way, I am happy.

4. kith says:

Quick Latex pointer: if you write "left(" and "right)" instead of "(" and ")" you get parentheses which self-adjust their height.