Convergence of double summation of divergent and convergent series I am looking at two series
(a)$$\sum_{n = 1}^\infty \dfrac{1}{n^2 (n + 1)^{2p}}\sum_{m = 1}^n (m + 1)^{2p}$$
and
(b)
$$\sum_{n = 1}^\infty \dfrac{1}{(n + 1)^{2p}}\sum_{m = 1}^n \dfrac{(m + 1)^{2p}}{m^2}$$
Assume $p \in (0, 1]$ for both (a) and (b). I wish to figure out which of the ones is summable and not summable (and for which power this holds).

Suppose $p = 1$ for simplicity, then, the first few terms of (a) are:
$\dfrac{2^2}{2^2} + \dfrac{2^2 + 3^2}{2^2 3^2} + \dfrac{2^2 + 3^2 + 4^2}{3^2 4^2} + \ldots = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{3^2 4} + \dfrac{1}{4^2} + \dfrac{1}{3^2} + \ldots$
Since this seems to converge at similar rate as $1/n^2$, does this indicate that (a) is convergent?

Similarly for (b), the first few terms are,
$\dfrac{2^2}{2^2} + \dfrac{2^2 + 3^2/2^2}{3^2} + \dfrac{2^2 + 3^2/2^2 + 4^2/3^2}{4^2} + \ldots =  1 + 2^2/3^2 + 1/2^2 + 1/4 + 3^2/4^3 + 1/3^2 + \ldots$
For this one I really have no idea what I could use to determine convergence.

The first series is divergent for all $p>0$, and therefore so is the second, whose terms are larger.
To see this, consider $\alpha>0$, and let's show that $\sum_{k=1}^nk^\alpha\sim Cn^{\alpha+1}$ for some constant $C$.
We have
$$\frac{1}{n^{\alpha+1}}\sum_{k=1}^nk^\alpha=\frac{1}{n}\sum_{k=1}^n\left(\frac kn\right)^\alpha$$
This is a Riemann sum, for the continuous function $f(x)=x^\alpha$, therefore
$$\frac{1}{n^{\alpha+1}}\sum_{k=1}^nk^\alpha\to\int_0^1 x^\alpha\mathrm dx=\frac{1}{\alpha+1}$$
Hence
$$\sum_{k=1}^nk^\alpha\sim\frac{n^{\alpha+1}}{\alpha+1}$$
Hence the term in the first series is equivalent to $\dfrac{n^{2p+1}}{(2p+1)n^2n^{2p}}\sim\dfrac{1}{(2p+1)n}$, and the series is divergent, per the limit comparison test.
For the second series, just note that
$$\sum_{m=1}^n\frac{(m+1)^{2p}}{m^2}\ge\frac{1}{n^2}\sum_{m=1}^n(m+1)^{2p}$$
