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

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 $a_k=\frac52 k^{3/2}+O(k^{1/2})$.


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)


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 $

Tags:

Sequences And Series

Real Analysis

Convergence Divergence

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