TITLE: Convergence of $\sum_{n=1}^{\infty}\left(\frac{1}{n^{2/3}}-\frac{7}{n^{3/2}}\right)$
QUESTION [0 upvotes]: I would please like your guidance to find if the series $$\sum_{n=1}^{\infty}\left(\frac{1}{n^{2/3}}-\frac{7}{n^{3/2}}\right)$$ converges or diverges?
I noticed that We have two separate $p$-series and $p<1$, so the conclusion is that the series is divergent?
Thank you very much in advance.

REPLY [0 votes]: You have been given many explanations about the divergence.
If you want more, consider
$$S_p=\sum_{n=1}^{p}\left(\frac{1}{n^{2/3}}-\frac{7}{n^{3/2}}\right)=H_p^{\left(\frac{2}{3}\right)}-7 H_p^{\left(\frac{3}{2}\right)}$$ where appear generalized harmonic numbers.
Using there asymptotics, you have
$$S_p=3 \sqrt[3]{p}+\left(\zeta \left(\frac{2}{3}\right)-7 \zeta
   \left(\frac{3}{2}\right)\right)+O\left(\frac{1}{\sqrt p}\right)$$ Using this truncated expression, you will see that $S_p > 0$ when $p > 331$ (in fact, this happens when $p > 292$) and then continues growing.
Using Wolfram Alpha for $p=10^4$ gives $44.040$ while the truncated expression given above leads to $43.899$.
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