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Show that $\displaystyle\sum_{n=1}^{\infty} (-1)^n \sin \left(\frac{x}{n}\right)$ converges uniformly on every finite interval.

At first thought, I try to find a bound, $M_n$, for the sine term such that $\displaystyle\sum_{n=1}^{\infty} M_n < \infty$.

However, the best I can find is that $\sin(\frac x n) \leq \frac R n + \frac{1}{n^3}$ for $x \in [-R,R]$

And now, I totally don't know how to proceed.

Thanks in advance.

PinkyWay
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Ashiroki
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  • The Taylor expansion of $\sin(y)$ gives $|\sin(\frac{x}{n})-\frac{x}{n}| < \frac{|x|}{n^3}$ for $|x/n| < \pi/2$ – reuns Nov 20 '16 at 22:18

2 Answers2

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Hint. One may recall that, by a Taylor series expansion, as $u \to 0$, we have $$ \sin u= u+O(u^3) $$ giving, as $n \to \infty$, $$ \sin \frac{x}{n}= \frac{x}{n}+O_x\left(\frac{1}{n^3}\right) $$ and, for any $N\ge1, M \ge1,$ $$ \sum_{n=N}^M(-1)^n\sin \frac{x}{n}=x \cdot \sum_{n=N}^M\frac{(-1)^n}{n}+\sum_{n=N}^M O_x\left(\frac{1}{n^3}\right) $$ one may deduce that $$ \left|\sum_{n=N}^M(-1)^n\sin \frac{x}{n}\right|\le |x| \cdot\left|\sum_{n=N}^M\frac{(-1)^n}{n}\right|+\sum_{n=N}^M\left|O_x\left(\frac{1}{n^3}\right)\right| $$ which yields the uniform convergence of the given series over each compact set $[-R,R]$, $R>0$.

Olivier Oloa
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  • I don't think this is right. You've shown the partial sums of the series are uniformly bounded, but that does not prove uniform convergence, or even pointwise convergence. For example, $|\sum_{n=1}^{N}(-1)^n|\le 1$ for all $N.$ – zhw. Nov 20 '16 at 20:08
  • @zhw You are right. But the path I've taken above works not only on partial sums but also on remainders, I've then edited my answer, thus the expected conclusion holds. Thank you for your feedback. – Olivier Oloa Nov 20 '16 at 22:11
  • You have shown the series is pointwise Cauchy, but $\mathcal{O}_x$ is not very clear. Why don't you assume $x \in [- R,R]$ at the beginning ? (and consider $n$ large enough, that is $|x/n| < \pi /2$) – reuns Nov 20 '16 at 22:20
  • Sure we have the pointwise convergence, but taking $\sup$ over $[-R,R]$ of both sides of the latter inequality, gives the uniform convergence on the given set. – Olivier Oloa Nov 20 '16 at 22:24
  • But you still have this $O_x$ business. You need a uniform $O(1/n^3).$ – zhw. Nov 20 '16 at 22:34
  • Please, are you unaware that this $\left|\mathcal{O_x}\right|$ is uniformly bounded over $|x| \leq R$? – Olivier Oloa Nov 20 '16 at 22:36
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Let $b>0.$ We'll show uniform convergence on $[-b,b].$ Because the terms of the series are odd, it's enough to prove uniform convergence on $[0,b].$

Let $0<\epsilon< \pi/2.$ Choose $N$ such that $b/N < \epsilon.$ Note that the sequence $\sin (x/N), \sin (x/(N+1)), \sin (x/(N+2)),\dots$ is then decreasing for each $x\in [0,b].$

Suppose $N\le m <n.$ Claim:

$$\tag 1 |\sum_{k=m}^{n}(-1)^n\sin (x/k)| <\epsilon \text { for } x\in [0,b].$$

Proof: The terms in this series decrease in absolute value and are alternating in sign. Therefore $(1)$ is no larger than $|\sin(x/m)|\le |x/m| \le b/m\le b/N < \epsilon.$ This proves the claim.

Note that the claim says precisely that the partial sums of our series are uniformly Cauchy on $[0,b].$ Hence the series converges uniformly on $[0,b].$

zhw.
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