# Convergence of series with modified denominator

Suppose the series with positive terms $\sum_{n=1}^{\infty} a_n$ converges. Let $r_n=\sum_{k=n}^{\infty}a_k$.

Prove or disprove that $\sum_{n=1}^{\infty}\frac{a_n}{r_n}$ diverges,

and prove or disprove that $\sum_{n=1}^{\infty}\frac{a_n}{\sqrt{r_n}}$ converges.

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## 1 Answer

1. $\sum_{n=1}^{\infty} a_n/r_n$ must diverge. For all $k$ and $\ell>k$, $$\sum_{k\le n\le \ell} \frac{a_n}{r_n}\ge \sum_{k\le n\le \ell} \frac{a_n}{r_k}= \frac{a_k+a_{k+1}+\dots+a_{\ell}}{a_k+a_{k+1}+\dots}.$$ The limit of the right-hand side as $\ell\to\infty$ is $1$, so for any $k$ we can choose $\ell>k$ such that $$\sum_{k\le n\le \ell} \frac{a_n}{r_n}>\frac12.$$ Therefore, the sum $\sum_n a_n/r_n$ must diverge.

2. $\sum_{n=1}^{\infty} a_n/\sqrt{r_n}$ must converge. The partial sum $\sum_{1\le n\le \ell} \frac{a_n}{\sqrt{r_n}}$ can be rewritten as \begin{eqnarray*} \sum_{1\le n\le \ell} \frac{r_n-r_{n+1}}{\sqrt{r_n}} &\le& \sum_{1\le n\le \ell} \ \int_{r_{n+1}}^{r_n} \frac{1}{\sqrt{x}} \, dx\\ &=& \int^{r_1}_{r_{\ell+1}} \frac{1}{\sqrt{x}} \, dx\\ &\le& \int_0^{r_1} \frac{1}{\sqrt{x}} \, dx\\ &=& 2\sqrt{r_1}. \end{eqnarray*} Therefore, the sum $\sum_n a_n/\sqrt{r_n}$ has bounded partial sums, and it is a sum of positive terms, so it converges. The same method of proof could be used to show that $\sum_n a_n \phi(r_n)$ converges for any positive function $\phi$ such that, for some $\epsilon>0$, $\phi$ is decreasing on $(0,\epsilon]$ and $\int_0^\epsilon \phi(x) \, dx <+\infty$.

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