Limit in combination with an infinite series

Question

How would I go about showing the following limits that involve infinite series

$$ \lim_{x \to 0^{+}} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{2k+1}} \sin (2\pi n(x - \frac{1}{2})) = 0 \text{ with } k \in \mathbb{N} $$

and

$$ \lim_{x \to 0^{+}} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{2k}} \cos (2\pi n(x - \frac{1}{2})) = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{2k}} \cos (-\pi n) = \sum_{n=1}^{\infty} \frac{1}{n^{2k}} \text{ with } k \in \mathbb{N} $$

I first had the idea that perhaps the rule of "the limit of the sums is the sum of the limits" would extend to the realm of infinite convergent series, but I came up with a counter example

$$ \lim_{x \to 0^{+}} \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sin (2\pi n(x - \frac{1}{2})) = \frac{\pi}{2} \neq 0 $$

I think I can prove the first limit by using Bernoulli polynomials, but that is not the kind of proof I'm looking for (I want to prove something about Bernoulli numbers by using this limit, so it would be kind of invalid to do so).

Idem so for limit no. 2, by using Euler's expression for even values of the Riemann zeta function, but again I want to prove something the other way around.

Any ideas? I must add that I'm an undergraduate math student, so my knowledge is very limited :-)

Answer 1

For $k=1,2,3,...$, you have $$ \begin{align} 0\leq\left|\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{2k+1}} \sin (2\pi n(x - \frac{1}{2}))\right| &\leq \sum_{n=1}^{\infty}\frac{1}{n^{2k+1}}=\zeta(2k+1)<\infty\\\\ 0\leq\left|\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{2k}} \cos (2\pi n(x - \frac{1}{2}))\right| &\leq \sum_{n=1}^{\infty}\frac{1}{n^{2k}}=\zeta(2k)<\infty \end{align} $$ thus, by uniform convergence on $[0,1]$, it is allowed to interchange the 'limit' and the 'infinite sum', then conclude with $$ \begin{align} \sin (2\pi n(x - \frac{1}{2})) \to 0 & \,\, \text{as}\,\, x \to 0^+,\\\\ (-1)^n\cos (2\pi n(x - \frac{1}{2})) \to 1 & \,\, \text{as}\,\, x \to 0^+. \end{align} $$

Answer 2

The funny thing is that the counter example you gave (k=0) was the ONLY valid one. Let us see why :

First, set $k > 0$. Then, $$ \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{2k+1}} \sin (2\pi n(x - \frac{1}{2})) = \sum_{n=1}^{\infty} \frac{1}{n^{2k+1}} \sin (2\pi nx) $$

Then, $ | \frac{1}{n^{2k+1}} \sin (2\pi nx) | \leq \frac{1}{n^{2k+1}} $ and $\sum \frac{1}{n^{2k+1}}$ converges. This means that your sum converges absolutly, hence you can exchange the limit and sum operations.

If $k=0$ on the other hand, the series does not absolutely converge and the values of the sum of the limits and the limit of the sums can be (and is) different.

Cheers,