Pointwise Convergence for a Series Given the series $$\sum_{n=1}^\infty \frac{\sin(nx)}{1+n^2x^2}$$
for $x$ in $\mathbb R$. Prove that the series converges pointwise for all $x$ in $\mathbb R$.
So I know for sequences, to show pointwise convergence you can just take the limit. Does this also work for series? Or is the pointwise convergence different for series?
 A: You can try and calculate the pointwise limit, which amounts to calculating 
$$ \lim_{N \to \infty} \sum_{n=1}^N \frac{\sin(nx)}{1 + n^2x^2} $$
for all $x \in \mathbb{R}$. However, as it often happens to be the case with series, you usually can't calculate the limit of a series but you can argue that it converges without actually knowing what it converges to by using various tests.
In your case, if we assume that $x \neq 0$, we have
$$ \sum_{n=1}^{\infty} \left| \frac{\sin(nx)}{1 + n^2x^2} \right| \leq \sum_{n=1}^{\infty} \frac{1}{1 + n^2 x^2} $$
and the positive series on the right hand side converges (for example, by comparing it to the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$). Thus, the original series $\sum_{n=1}^{\infty} \frac{\sin(nx)}{1 + n^2x^2}$ converges absolutely and in particular, converges for all $x \neq 0$. I'll leave the case $x = 0$ to you.
A: A standard theorem says that if $\displaystyle \sum_{n=1}^\infty |a_n|<\infty$ then $\displaystyle\sum_{n=1}^\infty a_n$ converges.
If $x\ne0$ then
\begin{align}
\sum_{n=1}^\infty \left|\frac{\sin(nx)}{1+n^2x^2} \right| & \le \sum_{n=1}^\infty \frac 1 {1+n^2x^2} & & \text{because } |\sin(\text{any real number})| \le 1 \\[10pt]
& \le \sum_{n=1}^\infty \frac 1 {n^2 x^2} & & (\text{recall that }x\ne 0) \\[10pt]
& = \frac 1 {x^2} \sum_{n=1}^\infty \frac 1 {n^2} & & \text{because $x^2$ does not depend on $n$} \\[10pt]
& < \infty.
\end{align}
The case where $x=0$ needs to be dealt with separately, but that is easy because $\sin(nx)=0$ in that case.
A: Yes, if $f_n : \Bbb R \to \Bbb R$ are real valued functions, then $f_n$ is said to converge pointwise to a function $f: \Bbb R \to \Bbb R$ if for any given $x \in \Bbb R$ there is a $N > 0$ for any $\epsilon > 0$ such that $|f_n(x) - f(x)| < \epsilon$ for all $n > N$. That is to say, for any fixed $x$, $\{f_n(x)\}$ converges as a sequence to $f(x)$.
In this case $f_n(x) = \displaystyle \sum_{k=1}^n \dfrac{\sin(nx)}{1+n^2x^2}$. All you have to do is to show that for any given $x \in \Bbb R$, $\{f_n(x)\}$ converges as a sequence as $n \to \infty$.
