# Insights Using the Fourier Series To Find Some Interesting Sums - Comments

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1. Dec 22, 2016

### Svein

2. Dec 25, 2016

### micromass

Staff Emeritus
3. Dec 26, 2016

### Svein

Yes - but I thought it best to keep it reasonably simple the first time through.

4. Jan 1, 2017

### stevendaryl

Staff Emeritus
Very nice. I have done similar tricks for evaluating summations, but I didn't know the trick of using Parseval’s formula. My favorite trick is using representations of the Dirac delta function, and that's how I would evaluate $\sum \frac{1}{n^2}$. However, it's a lot more convoluted.

$\delta(x) = \frac{1}{2\pi} + \frac{1}{\pi} \sum_n cos(nx)$

Now, integrate both sides from $-x$ to $+x$.

$sign(x) = \frac{x}{\pi} + \frac{2}{\pi} \sum_n \frac{1}{n} sin(nx)$

(where $sign(x) = \pm 1$ depending on whether $x>0$ or $x < 0$)

Integrate again, this time from $0$ to $x$:

$|x| = \frac{x^2}{2\pi} + \frac{2}{\pi} \sum_n \frac{1}{n^2} (1 - cos(nx))$

Using a trig identity, $1-cos(nx) = 2 sin^2(\frac{n}{2} x)$. So we have:

$|x| = \frac{x^2}{2\pi} + \frac{4}{\pi} \sum_n \frac{1}{n^2} sin^2(\frac{n}{2} x)$

Now, we set $x = \pi$ to get the identity:

$\pi = \frac{\pi}{2} + \frac{4}{\pi} \sum_n \frac{1}{n^2} sin^2(\frac{n}{2} \pi) = \frac{4}{\pi} \sum_{odd\ n} \frac{1}{n^2}$

So:

$\frac{\pi}{2} = \frac{4}{\pi} \sum_n \frac{1}{n^2} sin^2(\frac{n}{2} \pi) = \frac{4}{\pi} \sum_{odd\ n} \frac{1}{n^2}$

Drat! The sum on the right side is only over odd values of $n$, because $sin^2(\frac{n}{2} \pi) = 0$ when $n$ is even. But all is not lost. We can reason as follows:

$\sum_n \frac{1}{n^2} = \sum_{odd\ n} \frac{1}{n^2} + \sum_{even\ n} \frac{1}{n^2}$

but $\sum_{even\ n} \frac{1}{n^2} = \sum_{n} \frac{1}{(2n)^2} = \frac{1}{4} \sum_{n} \frac{1}{n^2}$. So we have:

$\sum_n \frac{1}{n^2} = \sum_{odd\ n} \frac{1}{n^2} + \frac{1}{4}\sum_{n} \frac{1}{n^2}$

So $\sum_{odd\ n} \frac{1}{n^2} = \frac{3}{4} \sum_n \frac{1}{n^2}$. Putting this back into our result, we have:

$\frac{\pi}{2} = \frac{4}{\pi} \sum_n \frac{1}{n^2} sin^2(\frac{n}{2} \pi) = \frac{4}{\pi} \frac{3}{4} \sum_{n} \frac{1}{n^2}$

So $\sum_n \frac{1}{n^2} = \frac{\pi^2}{6}$

Using Parseval is a lot simpler.

5. Feb 2, 2017