Examine the continuity and differentiability of functions:
a) $\displaystyle f(x)=\sum_{n=1}^{+\infty}\frac{\sin(nx)}{n^3}$
b) $\displaystyle f(x)=\sum_{n=1}^{+\infty}\arctan\left(\frac{x}{n^2} \right)$
in the case of differentiability explore the sign of $f '(0)$.
So, we are dealing with function series I think. I tried a):
Let $f_n(x)=\frac{\sin(nx)}{n^3}$. With Weierstrass M-test $|f_n|\le \frac{1}{n^3}$ so the series $\sum_{n=1}^{+\infty}f_n$ converges uniformly. The same applies for series of derivatives that is: $\sum_{n=1}^{+\infty}\frac{\cos(nx)}{n^2}$ so function $f$ is differentiable and so it is continuous. $f'(0)>0$
Is this argumentation ok?
I don't know how to approach b).
Best Answer
Your argument for part a) is correct.
For b), think about where the problem occurs when we try to do similar steps. We can't prove uniform convergence immediately, because $x$ could be unbounded. So instead, for now just consider $x\in (-a,a).$ Then since $|\arctan x| \leq |x| $ we have $$\biggr|\sum_{n=1}^{\infty} \arctan \left( \frac{x}{n^2} \right) \biggr| \leq \sum_{n=1}^{\infty} \frac{a}{n^2} < \infty. $$
Thus the sum converges uniformly in $(-a,a).$
Now, the sequence of term by term derivatives converges uniformly in $(-a,a)$ as well since (as sos440 mentioned above) $$ \frac{d}{dx} \arctan \left( \frac{x}{n^2} \right) \leq \frac{1}{n^2}.$$
Thus, we conclude the sum is differentiable for all $x\in (a,-a).$ Since $a>0$ was arbitrary, the sum is differentiable for all $x\in \mathbb{R},$ with derivative $$ f'(x) = \sum_{n=1}^{\infty} \frac{1}{n^2 + (x/n)^2} .$$ In particular, $f'(0) = \pi^2/6 > 0.$