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Let $\phi(x)=\begin{cases}0, & 0\lt x\lt 1\\ 1, & 1\lt x\lt3 \end{cases}$

We have that the Fourier cosine series is given by $$\phi(x)=\begin{cases}0, & 0\lt x\lt1\\ \frac{4}{3}+\displaystyle\sum_{m=1}^{\infty}\frac{-2\sin\frac{m\pi}{3}}{m\pi}\cos\frac{m\pi x}{3}, & 1\lt x\lt3 \end{cases}$$

Put $x=0$ to find the sum

$\displaystyle 1+\frac{1}{2}-\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{8}-\frac{1}{10}-\frac{1}{11}+\cdots$

I tried the following

$$\phi(0)=\frac{4}{3}+\sum_{m=1}^{\infty}\frac{-2\sin\frac{m\pi}{3}}{m\pi}\\=\frac43-2\frac{\sin\frac{\pi}{3}}{\pi}-\frac{\sin\frac{2\pi}{3}}{\pi}-2\frac{\sin\pi}{3\pi}-\frac{\sin\frac{4\pi}{3}}{2\pi}-\cdots\\=\frac{4}{3}-\frac{\sqrt3}{\pi}-\frac{\sqrt3}{2\pi}-0+\frac{\sqrt3}{4\pi}\dots=\frac{4}{3}-\frac{\sqrt3}{\pi}(1+\frac{1}{2}-\frac{1}{4}\dots)=\ ? $$

And I'm stuck here,

What can I do here?

I greatly appreciate any assistance you may provide.

user441848
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2 Answers2

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The key observation, which will be exploited below in computing the series $$ S=\sum_{n=1}^\infty\frac{\sigma_n}{n}, $$ where $\sigma_n$ is repeating sequence of $(1,1,0,-1,-1,0)$, is that $$ \sigma_n=\frac{2}{\sqrt3}\sin\frac{\pi n}{3}. $$ It follows: $$ S=\frac{2}{\sqrt3}S',\text{ with } S'=\sum_{n=1}^\infty\frac{\sin\frac{\pi n}{3}}{n}.\tag{1} $$ It appears to be simpler to compute directly $S'$. Two approaches to perform this are given below.

The first way, suggested in question, exploits the Fourier series of the even periodic function: $$ \phi(x)=\begin{cases} 0& -1<x<1\\ 1& -3<x<-1\text{ or } 1<x<3 \end{cases};\quad\phi(x+6)=\phi(x), $$ which is: $$ \phi(x)=\frac{2}{3}-\sum_{n=1}^\infty\frac{2\sin\frac{\pi n}{3}}{\pi n}\cos\frac{\pi n x}{3}.\tag{2} $$ Substituting in (2) $x=0$ one obtains: $$ 0=\phi(0)=\frac{2}{3}-\sum_{n=1}^\infty\frac{2\sin\frac{\pi n}{3}}{\pi n}= \frac{2}{3}-\frac{2}{\pi}S'\Rightarrow S'=\frac{\pi}{3}. $$

The second way is more direct as it does not rely on a foreknowledge of an appropriate function for Fourier series. Observe that for $|x|<1$: $$ \sum_{n=0}^\infty{x^n\sin\frac{\pi n}{3}}=\frac{1}{2i}\left(\frac{1}{1-xe^{\frac{\pi i}{3}}}-\frac{1}{1-xe^{-\frac{\pi i}{3}}}\right) =\frac{x\sin\frac{\pi}{3}}{1-2x\cos\frac{\pi}{3}+x^2}=\frac{\sqrt{3}}{2}\frac{x}{1-x+x^2}. $$

Thus, $$ S'=\sum_{n=1}^\infty\frac{\sin\frac{\pi n}{3}}{n}=\sum_{n=1}^\infty\sin\frac{\pi n}{3}\int_0^1 x^{n-1}dx=\frac{\sqrt3}{2}\int_0^1\frac{dx}{1-x+x^2}=\frac{\pi}{3}. $$

Finally the sum in question is computed using (1) as: $$ S=\frac{2}{\sqrt3}S'=\frac{2\pi}{3\sqrt3}. $$

user
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  • How did you find that $S=\displaystyle \frac{2}{\sqrt3}\sum_{n=1}^{\infty} \frac{\sin\frac{\pi n}{3}}{n}$ ? – user441848 Apr 30 '18 at 19:17
  • @Alt. As $\sin\frac{\pi n}{3}$ is a repeating sequence of $0,\frac{\sqrt3}{2},\frac{\sqrt3}{2},0,-\frac{\sqrt3}{2},-\frac{\sqrt3}{2}$ the expression $\frac{2}{\sqrt3}\sin\frac{\pi n}{3}$ takes on the values 0,1,1,0,-1,-1, i.e. exactly the values which enter the sum in question. – user Apr 30 '18 at 19:55
  • The sum in question $\displaystyle 1+\frac{1}{2}-\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{8}-\frac{1}{10}-\frac{1}{11}+\cdots$ takes vaules 0,1,1,0,-1,-1,... ? How do you know? – user441848 Apr 30 '18 at 21:28
  • wait, I think you mean this $\phi(x)=\frac{2}{3}-\sum_{n=1}^\infty\frac{2\sin\frac{\pi n}{3}}{\pi n}\cos\frac{\pi n x}{3}$ as sum in question ? – user441848 Apr 30 '18 at 21:34
  • @Alt. Look at the numerators of the series you have written Do you see the pattern? Imagine that the series contains also 0/3, 0/6 and so on... – user Apr 30 '18 at 21:37
  • yes, it's $0,\sqrt3,\sqrt3,0,-\sqrt3,-\sqrt3$. Ok – user441848 Apr 30 '18 at 21:40
  • @Alt. Exactly. And it is the same as $\sqrt3(0,1,1,0,-1,-1)$. – user Apr 30 '18 at 21:45
  • oh I see, thanks. Looking again at your answer I think there's a small typo in $\phi(0)=0$ ? Forget about it, there's no typo. There's a question, why did you change to the interval -1<x<1 in the definition of $\phi(x)$? – user441848 Apr 30 '18 at 21:52
  • @Alt. I did it to ensure that $\phi(x)$ is an even function of $x$. Otherwise $\sin$-terms were needed in the Fourier series. – user Apr 30 '18 at 21:58
  • $?$ What do you mean with 'Otherwise sin-terms were needed in the Fourier series' ? – user441848 Apr 30 '18 at 22:07
  • @Alt. If a function is not even, its Fourier series necessary contains $\sin$ terms. And the series in your question contained only $\cos$-terms, contrary to given definition of $\phi(x)$ (neither you did mention that the function is periodic). This was the reason of my question in the comment to your question. – user Apr 30 '18 at 22:11
  • oh ok I think I am understanding, so I think the original definition of $\phi(x)$ given in the exercise is wrong because $\cos(x)$ is always even, but the interval $(0,1)$ didn't satisfy it – user441848 Apr 30 '18 at 22:21
  • It should be $4/3$ and not $2/3$ as you have in your definition of $\phi(x).$ $A_0=(2/3)\int_1^3(1)cos(0)dx=4/3$ – user441848 May 01 '18 at 00:05
  • $F [f (x)]=A_0\color {red}{/2}+\sum_{n=1}^\infty A_n\cos\frac {2\pi n}{T}x+B_n \sin\frac {2\pi n}{T}x$. – user May 01 '18 at 19:02
  • :) right, I forgot about it.. – user441848 May 01 '18 at 19:09
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In this answer, it is shown that $$ \sum_{k=-\infty}^\infty\frac{(-1)^k}{z+k}=\pi\csc(\pi z) $$ Therefore, $$ \begin{align} \sum_{k=0}^\infty(-1)^k\left(\frac1{3k+1}+\frac1{3k+2}\right) &=\sum_{k=0}^\infty(-1)^k\left(\frac1{3k+1}-\frac1{-3(k+1)+1}\right)\\ &=\sum_{k=0}^\infty\left(\frac{(-1)^k}{3k+1}+\frac{(-1)^{-k-1}}{3(-k-1)+1}\right)\\ &=\sum_{k=-\infty}^\infty\frac{(-1)^k}{3k+1}\\ &=\frac13\sum_{k=-\infty}^\infty\frac{(-1)^k}{k+\frac13}\\ &=\frac\pi3\csc\left(\frac\pi3\right)\\[6pt] &=\frac{2\pi}{3\sqrt3} \end{align} $$

robjohn
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