I am trying to show that for $p\geq1$ $$\limsup_{k\to\infty}\frac{\sqrt{k}\sum_{n=k}^\infty n^{-3p}}{\left(\sum_{n=k}^\infty n^{-2p}\right)^{3/2}}<\infty.$$ Some numerical calculations suggest that this is true. However, I am having trouble proving it.
We know that $c_1:=\sum_{n=1}^\infty n^{-2p} =\zeta(2p) <\infty$ and $c_2:=\sum_{n=1}^\infty n^{-3p}=\zeta(3p) <\infty$. I tried rewriting it as $$\frac{\sqrt{k}\sum_{n=k}^\infty n^{-3p}}{\left(\sum_{n=k}^\infty n^{-2p}\right)^{3/2}}= \frac{\sqrt{k}\left(c_1-\sum_{n=1}^{k-1} n^{-3p}\right)}{\left(c_2-\sum_{n=1}^{k-1} n^{-2p}\right)^{3/2}}$$
and then finding an upper bound for the numerator and a lower bound for the denominator separately, but the ones I tried are always too large/small.
Any hint would be appreciated.
Not only $\limsup\limits_{k\to\infty}$ but even $\lim\limits_{k\to\infty}$ exists. For $\alpha>1$ we have $$\lim_{k\to\infty}k^{\alpha-1}\sum_{n=k}^{\infty}n^{-\alpha}=\lim_{k\to\infty}\frac1k\sum_{n=k}^{\infty}\Big(\frac{n}{k}\Big)^{-\alpha}=\int_1^\infty x^{-\alpha}\,dx=\frac{1}{\alpha-1},$$ thus $$\lim_{k\to\infty}\frac{\sqrt{k}\sum_{n=k}^{\infty}n^{-3p}}{\left(\sum_{n=k}^{\infty}n^{-2p}\right)^{3/2}}=\lim_{k\to\infty}\frac{k^{3p-1}\sum_{n=k}^{\infty}n^{-3p}}{\left(k^{2p-1}\sum_{n=k}^{\infty}n^{-2p}\right)^{3/2}}=\frac{(2p-1)^{3/2}}{3p-1}.$$