I saw this question posted from 2 years ago, but I don't quite understand the accepted solution using the Reimann series. Can someone explain how to solve this?

Show that

$f(x)=\sum_{k=1}^\infty \frac{1}{k}\sin\left(\frac{x}{k+1}\right)$

converges, point-wise on $R$ and uniformly on each bounded interval in $R$, to a differentiable function $f$ which satisfies

$|f(x)|\leq |x| \text{ and } |f'(x)|\leq 1$

for all $x$ in $R$.

Here is the original post:

Show that $f(x)=\sum_{k=1}^\infty \frac{1}{k}\sin(\frac{x}{k+1})$ converges.

I'm not sure what this notation means or how it proves point-wise convergence:

$f_n(x)\sim\frac{x}{n^2},\forall x\neq0$

  • $\begingroup$ It's difficult to answer questions of the form "I don't understand this (rather long) proof" without more assistance. Can you elaborate on which parts, specifically, in that explanation you do not understand? $\endgroup$ – Dan Feb 24 '16 at 22:30
  • $\begingroup$ Yeah, I edited my question. $\endgroup$ – Pareod Feb 24 '16 at 22:33
  • $\begingroup$ Thanks! I can answer that! :) $\endgroup$ – Dan Feb 24 '16 at 22:35
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    $\begingroup$ Possible duplicate of Show that $f(x)=\sum_{k=1}^\infty \frac{1}{k}\sin(\frac{x}{k+1})$ converges. $\endgroup$ – user147263 Feb 24 '16 at 23:36
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    $\begingroup$ What do you mean possible duplicate? I explicitly stated that I'm bringing this question back up to clarify and explore new solutions, and I linked the original post. $\endgroup$ – Pareod Feb 25 '16 at 3:24

Use Taylor series for sine:

$$\left|\frac1k\sin\frac x{k+1}\right|=\left|\frac1k\left(\frac x{k+1}+\mathcal O\left(\frac{x^3}{(k+1)^3}\right)\right)\right|\le$$

$$\le \frac{|x|}{k(k+1)}+\mathcal O\left(\frac{|x|^3}{k(x+1)^3}\right)$$

Observe the last term's series converges for any $\;x\in\Bbb R\;$ .


As in the accepted answer to that question, set

$$ f_n(x) = \frac 1 n \sin\left(\frac x {n+1}\right). $$

Presumably, what the author meant by $f_n(x) \sim \frac x {n^2}\,\forall x \neq 0$ is that $f_n(x)$ is approximately $x/n^2$ at each $x \neq 0$ (and, of course, at $x = 0$ as well, but he's handled that case separately). This follows from the formula that $\sin\theta$ is approximated by $\theta$ for $\theta$ near $0$, which you can infer, for instance, from the Maclaurin series of sine. Actually, we just need the Mean Value Theorem here, for it implies that for all $x$ there exists $c$ between $0$ and $x$ such that

$$ |f_n(x)| \leq |f_n(0)| + |f_n'(c)\cdot x| = 0 + \frac{|\cos(c/(n+1))|}{n(n+1)}|x| \leq \frac{|x|}{n(n+1)} \leq \frac{|x|}{n^2}. $$

What he does next (using normal convergence to get pointwise convergence) works but is unnecessary. We need only note that the above estimate implies for all $N$ and $x$ the partial sum $S_N(x) := \sum_{n=1}^\infty f_n(x)$ satisfies

$$ |S_N(x)| \leq |x|\sum_{n=1}^N \frac{1}{n^2}, $$

and hence $f(x) = \lim_{N\to\infty}S_N(x)$ exists and is bounded above by $|x|\sum_{n=1}^\infty \frac{1}{n^2}$.


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