2
$\begingroup$

It is given that $f(x)=|x|$ in $-\pi\le x\le \pi$ then the value of $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$$ is equal to

a) $\Large \frac{\pi^2}{2}\qquad$ b) $\Large\frac{\pi^2}{4}\qquad$ c) $\Large\frac{\pi^2}{6}\qquad$ d) $\Large\frac{\pi^2}{8}$

I expanded the series $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$$ $$=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\ldots$$ Now, I don't have any clue how to proceed.

$\endgroup$

marked as duplicate by Did, Martin Sleziak, Behrouz Maleki, R_D, GEdgar Sep 27 '16 at 17:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 6
    $\begingroup$ And what does this have to do with $f(x)=|x|$? Are you expected to use Fourier theory to arrive at your result? $\endgroup$ – Hagen von Eitzen Sep 27 '16 at 8:39
  • $\begingroup$ See this question: math.stackexchange.com/q/646548 $\endgroup$ – polfosol Sep 27 '16 at 8:56
9
$\begingroup$

Hint: $$ \sum_{n=1}^\infty \frac1{n^2} = \sum_{n=1}^\infty \frac1{(2n)^2}+\sum_{n=1}^\infty \frac1{(2n-1)^2}.$$

$\endgroup$
  • $\begingroup$ @ Vincenzo how does it help?. please explain me more $\endgroup$ – Bhaskara-III Sep 27 '16 at 8:48
  • $\begingroup$ It helps if you know that $\sum_{n=1}^{+\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$. If not, I suggest you use Fourier theory as suggested in the comments. $\endgroup$ – Augustin Sep 27 '16 at 8:51
  • 1
    $\begingroup$ @Bhaskara: I guess you know the value of the leftmost sum. See if you can "simplify" something on the right-hand side and then try isolating your sum. $\endgroup$ – Vincenzo Oliva Sep 27 '16 at 8:52
8
$\begingroup$

If we compute the Fourier cosine series of $f(x)$ over $(-\pi,\pi)$ through $$\forall n\geq 1,\qquad \int_{-\pi}^{\pi}f(x)\cos(nx) = 2\int_{0}^{\pi}x\cos(nx)\,dx = 2\,\frac{-1+\cos(\pi n)}{n^2}\tag{1}$$ we get: $$ |x|=\frac{\pi}{2}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos((2n-1)x)}{(2n-1)^2} \tag{2} $$ and by evaluating both sides at $x=0$ (pointwise convergence - as a by-product of uniform convergence - is ensured by the fact that the Fourier coefficients give a summable sequence) $$ \sum_{n\geq 1}\frac{1}{(2n-1)^2}=\color{red}{\frac{\pi^2}{8}}\tag{3} $$ follows. On the other hand, it is trivial that $$ \sum_{n\geq 1}\frac{1}{(2n-1)^2}\leq 1+\int_{1}^{+\infty}\frac{dx}{(2x-1)^2}=\frac{3}{2} \tag{4}$$ hence $(d)$ is the only reasonable option.

$\endgroup$
  • $\begingroup$ @Jack : how did you get the result of (2) i.e. $|x|=\frac{\pi}{2}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos((2n-1)x)}{(2n-1)^2}$ ? $\endgroup$ – Bhaskara-III Sep 27 '16 at 11:35
  • $\begingroup$ @Bhaskara-III: how do you compute a Fourier series, usually? I just did it. Notice that the coefficient $c_0$ is given by $$\frac{1}{2\pi}\int_{-\pi}^{\pi}|x|\,dx = \frac{\pi}{2}.$$ $\endgroup$ – Jack D'Aurizio Sep 27 '16 at 11:53
  • $\begingroup$ i don't know please explain me how to compute a fourier series $\endgroup$ – Bhaskara-III Sep 27 '16 at 11:54
  • $\begingroup$ @Bhaskara-III: have a look at here: en.wikipedia.org/wiki/Fourier_series $\endgroup$ – Jack D'Aurizio Sep 27 '16 at 11:55
  • $\begingroup$ Why does the question mention $f(x)$? I can probably use some other function to compute this sum. $\endgroup$ – ps95 Sep 27 '16 at 15:12
0
$\begingroup$

Use the Fourier Series of the function $f$ in the interval $[-\pi, \pi]$ And substitute $x = 0$. The series converges at this point to the value of $f$. This gives you the desired result.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.