Evaluating $\sum_{k=nq}^{np-1}$... I need to show that those two limits exist and state a formula for them:
a) $$\lim_{n\to\infty} \sum_{k=n}^{2n-1}\frac{n}{k^2}$$
b) $$\lim_{n\to\infty} \sum_{k=nq}^{np-1}\frac{1}{k}$$
with $p,q \in \mathbb{N}, q<p$. I don't know how to approach the problem with the sum starting at $n$.
 A: You have that
$$s_n:=\sum_{k=n}^{2n-1}\frac{n}{k^2} \le \sum_{k=n}^{2n-1} \frac{1}{n} = \frac{2n-n}{n} = 1\,,$$
hence the sequence $s_n$ is bounded from above. Now consider
$$s_{n+1}-s_n = \sum_{k=n+1}^{2n+1}\frac{n}{k^2}-\sum_{k=n}^{2n-1}\frac{n}{k^2}=\frac{n}{(2n+1)^2}+\frac{n}{(2n)^2}-\frac{1}{n} = -\frac{1+4n}{4n(2n+1)^2} \le0\,,$$
hence it is monotonically decreasing. It follows that the limit exists.
We can provide a lower bound similarly to how we got an upper bound
$$s_n:=\sum_{k=n}^{2n-1}\frac{n}{k^2} \ge \sum_{k=n}^{2n-1} \frac{n}{(2n-1)^2} = \frac{n^2}{(2n-1)^2} \ge \frac 14\,,$$

Regarding the closed formula, I think it involves the Gamma function (or its derivative, maybe). Definitely not a `nice' one.
EDIT: just realized you asked for the limit, not the closed formula. You could compute it through the Riemannian sums. Relabel your indexes so that
$$\sum_{k=n}^{2n-1}\frac{n}{k^2} = \frac{1}{n}\sum_{k=0}^{n-1}\frac{n^2}{(k+n)^2} = \frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\left(\frac{k+n}{n}\right)^2}=\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\left(1 + \frac{k}{n}\right)^2} = \int_0^1 \frac{dx}{(1+x)^2} = \frac 12$$

You could proceed the same for the other series.
