Find value of $p$ to make the series $\sum\limits_{n=1}^\infty\left(\dfrac1{n^p}\sum\limits_{k=1}^nk^{3/2}\right)$ converge 
Find value of $p$ that makes $\sum\limits_{n=1}^\infty\left(\dfrac1{n^p}\sum\limits_{k=1}^nk^{3/2}\right)$ converge.

I am using comparaison test: $0<k^{\frac{3}{2}}<k^2$, $0<\sum{k^{\frac{3}{2}}}<\sum k^2$. Since $\sum k^2=\dfrac{1}{6}n(n+1)(2n+1)$, then,
$$\sum _ { n = 1 } ^ { \infty } ( \frac { 1 } { n ^ { p } } \sum _ { k = 1 } ^ { n } k ^ { 3 / 2 } )<\sum _ {n=1} ^{\infty} \frac{1}{n^{p-3}}$$
By p-series $$\sum _ {n=1} ^{\infty} \frac{1}{n^{p-3}}$$ Converges when $$p>4$$
Have other test and how about my method right or wrong ?
 A: 
I thought it might be instructive to present an approach that relies only on creative telescoping and expansion of $\displaystyle \left(1-\frac1k\right)^{5/2}$ only, and avoids appealing to integrals.  To that end, we proceed.


Let $a_k=k^{5/2}-(k-1)^{5/2}$.  Then, we have
$$\sum_{k=1}^n a_k=n^{5/2}\tag1$$

Next, expanding $a_k$, we find that
$$\begin{align}
a_k&=k^{5/2}-(k-1)^{5/2}\\\\
&=k^{5/2}\left(1-\left(1-\frac1k\right)^{5/2}\right)\\\\
&=\frac52 k^{3/2}+O\left(k^{1/2}\right)\tag2
\end{align}$$

Putting together $(1)$ and $(2)$ reveals
$$\sum_{k=1}^n k^{3/2}=\frac25 n^{5/2}+\sum_{k=1}^n O\left(k^{1/2}\right)$$

So, in order for the series $\sum_{n=1}^\infty n^{-p}\sum_{k=1}^n k^{3/2}$ to converge $p-5/2>1$ or $p>7/2$.
And we are done!

APPENDIX:
We can expand $a_k=k^{5/2}-(k-1)^{5/2}$ without use of calculus as follows.  We write
$$\begin{align}
a_k&=k^{5/2}-(k-1)^{5/2}\\\\
&=\left(k^{1/2}-(k-1)^{1/2}\right)\left(k^2+k^{3/2}(k-1)^{1/2}+k(k-1)+k^{1/2}(k-1)^{3/2}+(k-1)^2\right)\\\\
&=\frac{k^2+k^{3/2}(k-1)^{1/2}+k(k-1)+k^{1/2}(k-1)^{3/2}+(k-1)^2}{k^{1/2}+(k-1)^{1/2}}\\\\
&=k^{3/2}\frac{1+(1-1/k)^{1/2}+(1-1/k)+(1-1/k)^{3/2}+(1-1/k)^2}{1+(1-1/k)^{1/2}}\\\\
&=\frac52 k^{3/2}\left(\frac25\,\frac{1+(1-1/k)^{1/2}+(1-1/k)+(1-1/k)^{3/2}+(1-1/k)^2}{1+(1-1/k)^{1/2}}\right)\tag{A1}
\end{align}$$
Note that the term in parentheses on the right-hand side of $(A1)$ goes to $1$ as $k\to \infty$.
So, $a_k=\frac52 k^{3/2}(1+o(1))$.  One can continue the expansion to show that in fact $a_k=\frac52 k^{3/2}+O(k^{1/2})$.
A: Using generalized harmonic numbers$$S_n=\sum\limits_{k=1}^nk^{3/2}=H_n^{\left(-{3/2}\right)}$$ the asymptotics of which being
$$H_n^{\left(-3/2\right)}=\frac{2 n^{5/2}}{5}+\frac{n^{3/2}}{2}+\frac{n^{1/2}}{8}+\zeta
   \left(-\frac{3}{2}\right)+O\left(\frac{1}{n^{3/2}}\right)$$ could help to have a better idea (almost if $p$ is not an integer)
A: I think it's a good idea to also share the integral method.
Let's $a=\dfrac{3}{2} $
$t \to t^a $ is an inscrasing function
So giving $(n,k)\in {\mathbb{N}^*}^2$
$$\int_{k-1}^{k} t^a dt \leq k^a \leq \int_{k}^{k+1}$$
Summing for $k$
$$\int_{0}^{n} t^a \leq \sum_{k=0}^n k^a \leq \int_{1}^{n+1} t^a $$
Which is :
$$\dfrac{1}{a+1} n^{a+1} \leq \sum_{k=0}^1 k^a \leq \dfrac{1}{a+1} (n+1)^{a+1}-1$$
Because $a+1 \geq 0$
Both left and right terms are equivalent to $$ \dfrac{1}{a+1} n^{a+1} $$
Hence you general term is equivalent to
$$\dfrac{1}{a+1} n^{a+1-p}$$
So we have convergence if and only if :
$$ p>a = \frac{7}{2}$$
I made the calculus with $a$ so you can repeat it for other values (for series) since it fits to the condition $ a+1>0 $
A: More generally,
if $a > 0$ then
$\sum_{k=1}^n k^a
\approx n^{a+1}/(a+1)
$
so 
$\sum_n \frac1{n^p}\sum_{k=1}^n k^a
\approx \sum_n \frac1{(a+1)n^{p-a-1}}
$ 
converges if
$p-a-1 > 1
$
or $p > a+2$
and diverges if
$p \le a+2$.
Here,
$a = 3/2$
so the sum
converges if
$p > 7/2$
and diverges if
$p \le 7/2$.
