Point-wise and Uniform Convergence 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:
$f_n(x)\sim\frac{x}{n^2},\forall x\neq0$
 A: 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 $\;x\in\Bbb R\;$ .
A: 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 $|x|\sum_{n=1}^\infty \frac{1}{n^2}$.
