On pointwise convergence of Fourier series

So, the function is piecewise continuous (and differentiable), with (generalized) one-sided derivatives existing at the points of discontinuity. Hence I conclude from the theorem that the series converges pointwise for all ##t## to the function ##f##.

I've double checked with WolframAlpha that the coefficients are \begin{align} c_0&=\frac{\pi}{4} \nonumber \\ c_{2n}&=\frac{i}{4n}, \ n\neq0 \nonumber \\ c_{2n+1}&=\frac{i}{2(2n+1)}+\frac{1}{\pi(2n+1)^2}. \nonumber \end{align} Now, we have ##\sum_{n\in\mathbb Z} c_ne^{int}##, and I proceed with the following computation \begin{align} \sum_{n\in\mathbb Z} c_ne^{int}&=c_0+\sum_{\substack{n\in\mathbb Z \\ n\neq0}} c_{2n}e^{i2nt}+\sum_{n\in\mathbb Z}c_{2n+1}e^{i(2n+1)t} \tag1 \\ &=\frac{\pi}{4}+\sum_{\substack{n\in\mathbb Z \\ n\neq0}} \frac{i}{4n}e^{i2nt}+\sum_{n\in\mathbb Z } \left(\frac{i}{2(2n+1)}+\frac{1}{\pi(2n+1)^2}\right)e^{i(2n+1)t} \nonumber \\ &=\frac{\pi}{4}-\sum_{n=1}^\infty\frac{\sin(2n)t}{2n}-\sum_{n=0}^\infty \frac{\sin(2n+1)t}{2n+1}+\frac{2}{\pi}\sum_{n=0}^\infty\frac{\cos(2n+1)t}{(2n+1)^2}\nonumber \\ &= \frac{\pi}{4}-\sum_{n=1}^\infty\frac{\sin nt}{n}+\frac{2}{\pi}\sum_{n=1}^\infty\frac{\cos(2n-1)t}{(2n-1)^2}\nonumber. \end{align} To evaluate the sum in the problem, I set ##t=0## and keep in mind that this is a discontinuous point. I get $$\frac12(f(0+)-f(0-))=\frac{\pi}{2}=\frac{\pi}{4}+\frac{2}{\pi}\sum_{k=1}^\infty \frac1{(2n-1)^2},$$ i.e. ##\sum_{k=1}^\infty \frac1{(2n-1)^2}=\frac{\pi^2}{8}##.

I'm doubting my own computation though. The equality in ##(1)##, when we split the sum into even and odd indexed sums, doesn't this require absolute convergence of the series?

What does you text tell you about how to take the double limit [tex]\sum_{n \in \mathbb{Z}} c_ne^{inx} =
\lim_{M \to \infty}\lim_{N \to \infty} \sum_{n=-M}^{N} c_{n}e^{inx}?[/tex] If you are to assume [itex]M = N[/itex] then this procedure is justified: all the necessary terms appear in the partial sum.

pasmith said:

What does you text tell you about how to take the double limit [tex]\sum_{n \in \mathbb{Z}} c_ne^{inx} =
\lim_{M \to \infty}\lim_{N \to \infty} \sum_{n=-M}^{N} c_{n}e^{inx}?[/tex] If you are to assume [itex]M = N[/itex] then this procedure is justified: all the necessary terms appear in the partial sum.

Yes, ##M=N## is the definition in my book. Are you saying that if the right-hand side converges (at least pointwise) in $$\sum_{n=-N}^{N}c_ne^{inx}=c_0+\sum_{\substack{n=-N \\ n \text{ even}}}^N c_{n}e^{inx}+\sum_{\substack{n=-N \\ n \text{ odd}}}^N c_{n}e^{inx},$$ then we can split the sum (technically rearrange it) into sums of even and odd indexed terms? I know it holds for positive series indexed by ##\mathbb N##, but I'm not sure if it holds for alternating, possibly complex, series indexed by ##\mathbb Z##.

If the answer is yes, I assume this follow from the limit law, where ##s_n=a_n+b_n## and ##\lim_{n\to\infty} a_n, \ \lim_{n\to\infty} b_n## both exist, then $$\lim_{n\to\infty} s_n=\lim_{n\to\infty} (a_n+b_n)=\lim_{n\to\infty} a_n+\lim_{n\to\infty} b_n.$$

psie said:

Yes, ##M=N## is the definition in my book. Are you saying that if the right-hand side converges (at least pointwise) in $$\sum_{n=-N}^{N}c_ne^{inx}=c_0+\sum_{\substack{n=-N \\ n \text{ even}}}^N c_{n}e^{inx}+\sum_{\substack{n=-N \\ n \text{ odd}}}^N c_{n}e^{inx},$$ then we can split the sum (technically rearrange it) into sums of even and odd indexed terms? I know it holds for positive series indexed by ##\mathbb N##, but I'm not sure if it holds for alternating, possibly complex, series indexed by ##\mathbb Z##.

You can't do [tex]
\lim_{N \to \infty} \sum_{n=-N}^N a_n = \lim_{N \to \infty} \sum_{\substack{n=-N \\ n \text{ odd}}}^N a_n + \lim_{N \to \infty} \sum_{\substack{n=-N \\ n \text{ even}}}^N a_n[/tex] because it involves taking all even terms before (or after) all odd terms, and that is only justified if the series is absolutely convergent. You can, of course, omit from the sum any term which is zero.

(Take care when indexing odd and even terms separately. You need the index to be symmetric about zero, so taking the even term as [itex]n = 2k[/itex] is fine since [itex]-2k = 2(-k)[/itex], but taking the odd term as [itex]n = 2k+1[/itex] is not since [itex]-(2k+1) \neq 2(-k) + 1[/itex]. The index [itex]n = (-1)^k(2|k| - 1)[/itex] for [itex]|k| \geq 1[/itex] does have this property.)

If the answer is yes, I assume this follow from the limit law, where ##s_n=a_n+b_n## and ##\lim_{n\to\infty} a_n, \ \lim_{n\to\infty} b_n## both exist, then $$\lim_{n\to\infty} s_n=\lim_{n\to\infty} (a_n+b_n)=\lim_{n\to\infty} a_n+\lim_{n\to\infty} b_n.$$

This theorem only applies if [itex]\lim a_n[/itex] and [itex]\lim b_n[/itex] exist separately; here they do not.

What you can do is [tex]
\lim_{N \to \infty} \sum_{n=-N}^N c_ne^{int} = c_0 + \lim_{N \to \infty}\sum_{n=1}^N ((c_n + c_{-n})\cos nt + i(c_n - c_{-n})\sin nt).[/tex] Now if [itex]f[/itex] is real then [itex]c_n[/itex] and [itex]c_{-n}[/itex] are complex conjugates, so this reduces to [tex]
c_0 + \lim_{N \to \infty} \sum_{n=1}^N (2\Re(c_n) \cos nt - 2\Im(c_n)\sin nt).[/tex] Setting [itex]t = 0[/itex] makes the sine terms vanish, and [itex]\Re(c_n)[/itex] vanishes for even [itex]n[/itex] so we are justified in writing this as [tex]
\tfrac12 (f(0^{+}) + f(0^{-})) = c_0 + \lim_{K \to \infty} \sum_{k=1}^K 2\Re(c_{2k - 1}).[/tex]