Does the following converge to 0? Does the following converge to $0$ for $\theta>-1/2$?
$$\lim_{n\rightarrow\infty}\frac{\sum_{i=1}^ni^{3\theta}}{\left(\sum_{i=1}^ni^{2\theta}\right)^{3/2}}$$
I'd like to use the comparison test but no idea what to compare it to. If $\theta=1$, based on code I wrote it seems to be going to 0 but would like to know how to use math.
 A: Rearranging terms we have that
$$\frac{\sum_{i=1}^ni^{3\theta}}{\left(\sum_{i=1}^ni^{2\theta}\right)^{\frac{3}{2}}}\cdot\frac{\frac{1}{n^{3\theta}}}{\frac{1}{n^{3\theta}}}\cdot\frac{\frac{1}{n^{\frac{3}{2}}}}{\frac{1}{n^{\frac{3}{2}}}} = \frac{\sum_{i=1}^n\left(\frac{i}{n}\right)^{3\theta}\frac{1}{n}}{\left(\sum_{i=1}^n\left(\frac{i}{n}\right)^{2\theta}\frac{1}{n}\right)^{\frac{3}{2}}}\cdot\frac{1}{\sqrt{n}}$$
The terms on the left converge to
$$\frac{\sum_{i=1}^n\left(\frac{i}{n}\right)^{3\theta}\frac{1}{n}}{\left(\sum_{i=1}^n\left(\frac{i}{n}\right)^{2\theta}\frac{1}{n}\right)^{\frac{3}{2}}} \longrightarrow\frac{\int_0^1x^{3\theta}dx}{\left(\int_0^1x^{2\theta}dx\right)^{\frac{3}{2}}} = \frac{(1+2\theta)^{\frac{3}{2}}}{1+3\theta}$$
if and only if $\theta>-\frac{1}{3}$, not just $-\frac{1}{2}$. Thus the overall sequence converges to $0$ in that case. If $-\frac{1}{2}<\theta\leq-\frac{1}{3}$, the denominator Riemann sum converges, and the numerator terms
$$N=\frac{1}{\sqrt{n}}\sum_{i=1}^n\left(\frac{i}{n}\right)^{3\theta}\frac{1}{n} = \frac{1}{\sqrt{n}}\sum_{i=1}^n\sqrt{\frac{n}{i}}\cdot\left(\frac{i}{n}\right)^{3\theta+\frac{1}{2}}\frac{1}{n}$$
$$\implies \frac{1}{\sqrt{n}}\sum_{i=1}^n\left(\frac{i}{n}\right)^{3\theta+\frac{1}{2}}\frac{1}{n} < N < \sum_{i=1}^n\left(\frac{i}{n}\right)^{3\theta+\frac{1}{2}}\frac{1}{n}$$
$$\longrightarrow 0 < N < \int_0^1x^{3\theta+\frac{1}{2}}dx = \frac{2}{6\theta+3}$$
by the same logic above. Thus the entire expression will always converge when $\theta > -\frac{1}{2}$, and only possibly to something nonzero when $-\frac{1}{2} < \theta \leq -\frac{1}{3}$.
A: Anothe solution using generalized harmonic nmbers
$$a_n=\frac{\sum_{i=1}^ni^{3\theta}}{\Big[\sum_{i=1}^ni^{2\theta}\Big]^{\frac 32}}=\frac{H_n^{(-3 \theta )}}{\Big[H_n^{(-2 \theta )}\Big]^{\frac 32}}$$ Using asymptotics
$$H_n^{(-3 \theta )}=n^{3 \theta } \left(\frac{n}{3 \theta +1}+\frac{1}{2}+\frac{\theta }{4 n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-3 \theta )$$
$$H_n^{(-2 \theta )}=n^{2 \theta } \left(\frac{n}{2 \theta +1}+\frac{1}{2}+\frac{\theta }{6 n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-2 \theta )$$
$$a_n\sim\frac{(2 \theta +1)^{3/2}}{3 \theta +1}\frac 1{\sqrt n}\Bigg[1-\frac1 {4n}+O\left(\frac{1}{n^2}\right)\Bigg]$$ and then ...
