# If $a_n>0$ and $\sum a_n$ converges then $\sum \frac{a_n}{\sqrt{r_n}}$ converges, where $r_n = \sum\limits_{m=n}^{\infty} a_m$

I've been stuck on the very last part of this problem from Baby Rudin for the past couple of days. If anyone could point me in the right direction, I'd appreciate it.

The Problem:

Suppose $a_n > 0$ and $\sum_{n=1}^{\infty} a_n$ converges. Put

$$r_n = \sum_{m=n}^{\infty} a_m .$$

Prove that

$$\frac{a_n}{\sqrt{r_n}} < 2(\sqrt{r_n} - \sqrt{r_{n+1}})$$

and deduce that $\sum \frac{a_n}{\sqrt{r_n}}$ converges.

Where I Am:

I've, indeed, proven the inequality; but, try as I might, I have not been able to prove (pardon me, "deduce") that $\sum \frac{a_n}{\sqrt{r_n}}$ converges. I've hit it with every inequality I can think of (so as to use the comparison test), and manipulated this thing algebraically every imaginable way, but it's been many days with no luck. Indeed, all I need to show is that

$$\frac{a_n}{\sqrt{r_n}} < 2(\sqrt{r_n} - \sqrt{r_{n+1}}) \le a_n$$

for all $n$. Something seemingly so quaint, so modest... and yet, this might be the one that breaks me.

• $$\sum_{n=1}^N\frac{a_n}{\sqrt{r_n}}\leqslant\sum_{n=1}^N2(\sqrt{r_n} - \sqrt{r_{n+1}})=2\sqrt{r_1}-2\sqrt{r_{N+1}}<2\sqrt{r_1}$$ – Did Oct 17 '17 at 6:28
• Are you also a math prof too? ..... – DeepSea Oct 17 '17 at 6:34
• – Martin R Oct 17 '17 at 6:41