1

How to find which $x\in[0,1]$ the sum $$\sum_{k=1}^\infty\frac{\cos(2\pi kx)}{k}$$ converge using the identity $$\sum_{k=1}^n a_kb_k=A_nb_n+\sum_{k=1}^{n-1}A_k(b_k-b_{k+1}), \qquad A_k=\sum_{j=1}^k a_j.$$

$x\in\{0,1\}$ clearly makes divergent as $$\sum_{k=1}^\infty\frac{1}{k}>\infty.$$

Using the identity $$\sum_{k=1}^n a_kb_k=A_nb_n+\sum_{k=1}^{n-1}A_k(b_k-b_{k+1}), \qquad A_k=\sum_{j=1}^k a_j,$$ we have \begin{align*}\sum_{k=1}^n\frac{\cos(2\pi kx)}{k}&=\sum_{k=1}^n\cos(2\pi kx)\cdot\frac{1}{k}\\&=\sum_{k=1}^n a_kb_k\\&=A_nb_n+\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})\\&=A_n\frac{1}{n}+\sum_{k=1}^{n-1}A_k\frac{1}{k(k+1)}.\end{align*}

Since $$\lvert A_k \rvert\le\sum_{j=1}^k \lvert a_j\rvert\le\sum_{j=1}^k 1=k$$ we have \begin{align*} \biggl|\sum_{k=1}^n\frac{\cos(2\pi kx)}{k}\biggr| &\le \lvert A_n\rvert \frac{1}{n}+\sum_{k=1}^{n-1}\lvert A_k\rvert \frac{1}{k(k+1)} \\&\le n\frac{1}{n}+\sum_{k=1}^{n-1}k \frac{1}{k(k+1)} \\&= 1+\sum_{k=1}^{n-1} \frac{1}{k+1}\end{align*} which does not help much.

Any suggestions how to make an upper bound that is more 'limited' so that I can find the $x$ for which it converges?

balddraz
  • 7,558
mf67
  • 853
  • 1
    I think $x \in (0,1)$ should make $|A_k|$ bounded (independent of $k$), or at least $\lim_{k \to \infty} A_k/k = 0$. For a proof of this, try to find an explicit formula for $\sum_{k=1}^K \cos(2\pi k x)$. – mathworker21 Jul 29 '19 at 19:49
  • Alternative solution here: https://math.stackexchange.com/q/815664/42969. – Martin R Jul 29 '19 at 20:02

1 Answers1

1

Fix $x\in(0,1)$. Using $$ \sin x\cos y=\frac{1}{2}\bigg[\sin(x+y)+\sin(x-y)\bigg], \sin x-\sin y=2\sin(\frac{x+y}2)\cos(\frac{x-y}2) $$ one has $$ \sin(\pi x)\sum_{k=1}^n\cos(2\pi kx)=\frac12\sum_{k=1}^n\bigg[\sin(\pi(2k+1)x)-\sin(\pi(2k-1)x)\bigg]=\frac12\bigg[\sin(\pi(2n+1)x)-\sin(\pi x)\bigg]=\sin(\pi(n+1) x)\cos(\pi n x) $$ and hence $$ \sum_{k=1}^n\cos(2\pi kx)=\frac{\sin(\pi(n+1) x)\cos(\pi n x)}{\sin(\pi x)}. $$ So $$ \bigg|\sum_{k=1}^n\cos(2\pi kx)\bigg|=\bigg|\frac{\sin(\pi(n+1) x)\cos(\pi n x)}{\sin(\pi x)}\bigg|\le\frac{1}{\sin(\pi x)}. $$

xpaul
  • 44,000
  • @PeterForeman, properly interpreted (in particular, changing from $\sum_{k=1}^n$ to $\sum_{j=1}^k$), this gives a more useful inequality than the OP's $|A_k|\le k$. – Barry Cipra Jul 29 '19 at 21:21
  • An interesting method indeed. Since the problem is from 'basic calculus' I don't think it is known that $\sum 1/k^2$ converge. I must think more on the identity how it can be used to solve the problem with very basic math knowledge. It's probably very easy, but currently I cannot see it… – mf67 Jul 29 '19 at 21:35
  • Ran through the calculations and a few typos in the @xpaul formulas, $2\sin(\tfrac{x-y}{2})\cos(\tfrac{x+y}{2})=\sin(x)-\sin(y)$ which gives $\sin(n\pi x)\cos\bigl((n+1)\pi x\bigr)$, otherwise it seems ok. – mf67 Jul 30 '19 at 09:04