Finding the infinite sum of a Fourier Series at a given $x$

Question

My question is this:

Let $f(x)=3x-1$, with period $1$ and $x\in(0,1)$.

Is the Fourier series of $f(x)$ convergent at $x=2/3$? If yes, what is the corresponding value of the sum of the Fourier series?

I thought that every Fourier series is convergent for all $x$, because the function is piecewise smooth and periodic, and we're defining the value of any jump discontinuities to be the $a_0$. Is that wrong? And if it's true, is it also true for a sine/cosine Fourier series? Or is it common in this topic to ask about uniform convergence and omit the "uniform"?

When it comes to computing the value, I get a little confused. I know by periodicity that $f(2/3)=f(-1/3)=-2$, and since $a_0=-1$ the infinite sum has to be convergent to $-1$, which I can easily verify on a computer program. Is periodicity sufficient to prove this is what the value of the Fourier series is?

If not, I cannot figure out how to find the sum by hand. I can get this far on my own: $f(2/3)=-2=-1 + \frac{3}{\pi}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}\sin(4\pi n/3)}{n} \Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^{n+1}\sin(4\pi n/3)}{n} = -\frac{\pi}{3}$, and I suspect I'm supposed to use some Calculus series information to somehow reach this conclusion. I've done quite a bit of research into the general form of different series, to try to refresh my memory, but none of them seem to be similar enough to my sum. Since there's a $\pi$ involved, I'm sure I can't just use the decimal approximation to see the partial sums are approaching that limit. There is probably a really simple way to do this, but I just can't figure it out:

If I write out the first several terms of the sum as in the last part above, I get $-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2\cdot2}+0+\frac{\sqrt{3}}{2\cdot4}+\frac{\sqrt{3}}{2\cdot5}-0-\frac{\sqrt{3}}{2\cdot7}-\frac{\sqrt{3}}{2\cdot8}+0+\frac{\sqrt{3}}{2\cdot10}+\frac{\sqrt{3}}{2\cdot11}-0...$ I thought maybe I should combine the positives and negatives so I could get back to a form with $(-1)^k$, and then I get $\frac{-3\sqrt{3}}{4}+\frac{9\sqrt{3}}{40}-\frac{15\sqrt{3}}{112}+\frac{21\sqrt{3}}{220}-...$, whose denominators are similar to the series $\sin^{-1}(x)\approx x+\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}+\frac{35x^9}{1152}+...$

If it helps, I do know that $\sin^{-1}(\frac{-\sqrt{3}}{2})=-\frac{\pi}{3},$ but I don't see how the given sum could be rewritten as the Taylor series for inverse sin. The denominators look the same except the last term, but if I plug in any $x$ to the Taylor series for $\sin^{-1}(x)$, I don't get an alternating sequence. And, if I try to use $\frac{-\pi}{3}=-\tan^{-1}(\sqrt{3}), \tan^{-1}(x) \approx x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ...$, since there's an alternating sequence, I get even further away from my equating the sum to a trig function.

If anyone could please explain to me how I should analyze sums like this, so that I can know what I'm doing enough to solve them by hand, and clarify the convergence question, I would greatly appreciate it!

Answer 1

The Fourier series converges to $f(x)$ at points of continuity, including $x=\frac{2}{3}.$

At points of discontinuity, $x =k\in \mathbb{Z}$ the series will converge to the average of the limits from the left and right:

$$ \textrm{Fourier series } \rightarrow \frac{1}{2} (\lim_{x\uparrow k} f(x) + \lim_{x \downarrow k} f(x)).$$

If the period is $T=1$, and the function is as shown in the figure, then the Fourier series is given by

$$f(x) \sim \frac{a_0}{2} + \sum_{k=1}^{\infty} (a_k \cos 2\pi k x + b_k \sin 2\pi k x).$$

$$a_0=2 \int_0^1 (3x-1 )\, dx = 1,$$

$$a_k=2 \int_0^{1} (3x-1) \cos 2\pi k x \, dx = 0,$$

$$b_k=2 \int_0^{1} (3x-1) \sin 2\pi k x \, dx = -\frac{3}{k \pi},$$

$$f(x) \sim \frac{1}{2} - \frac{3}{\pi} \sum_{k=1}^{\infty}\frac{\sin 2\pi k x}{k} .$$

$$f(2/3) \sim \frac{1}{2} + \frac{3\sqrt{3}}{2\pi} \sum_{k=0}^{\infty}\left[\frac{1}{3k+1} - \frac{1}{3k+2}\right]$$

$$f(2/3) \sim \frac{1}{2} + \frac{3\sqrt{3}}{2\pi} \sum_{k=0}^{\infty} \frac{1}{9k^2+9k+2} = \frac{1}{2}+\frac{1}{2} =1.$$

One way to evaluate the summation formula:

$$\sum_{k=0}^\infty \frac{1}{9k^2+9k+2} = \sum_{k=0}^\infty \int_1^2 \frac{1}{(3k+x)^2} dx = \int_1^2 \sum_{k=0}^\infty \frac{1}{(3k+x)^2} dx$$

$$= \frac{1}{9} \int_1^2 \psi_1 (x/3)dx = \frac{1}{3} [\psi_0(2/3)-\psi_0(1/3)] = \frac{\pi}{3\sqrt{3}}.$$

The function $\psi_n(\cdot)$, is the polygamma function.