# Baby Rudin chapter 3 problem 12.

Suppose $a_n > 0$ and $\sum a_n$ converges. Put $r_n = \sum_{m=n}^{\infty} a_m$.

1) Show that $\frac{a_m}{r_m} + ... + \frac{a_n}{r_n} > 1 - \frac{r_n}{r_m}$, if $m < n$, and deduce that $\sum \frac{a_n}{r_n}$ diverges.

2) Show 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.

I am stumped on this problem, do not know how to start. Any help would be great.

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Presumably the very last summation is missing a square root symbol? –  Gerry Myerson Jul 6 '12 at 4:59
Oops..thanks Gerry! –  Ellen Jul 6 '12 at 5:03
I think you do not want the square root in the sum appearing in 1). –  David Mitra Jul 6 '12 at 5:40
@David..thanks I fixed it/ –  Ellen Jul 6 '12 at 20:37

For problem 1, note that $$\frac{a_i}{r_i}=\frac{r_i-r_{i+1}}{r_i}$$ which gives us $$\frac{a_m}{r_m} + ... + \frac{a_n}{r_n}=\frac{r_m-r_{m+1}}{r_m}+\cdots+\frac{r_n-r_{n+1}}{r_n}>\frac{r_m-r_{m+1}+\cdots+r_n-r_{n+1}}{r_m}$$ and nice things happen when you cancel terms in the numerator.
For problem 2, note that $$\frac{a_n}{\sqrt{r_n}}=\frac{r_n-r_{n+1}}{\sqrt{r_n}}=\frac{r_n}{\sqrt{r_n}}-\frac{\sqrt{r_{n+1}}}{\sqrt{r_{n}}}\frac{r_{n+1}}{\sqrt{r_{n+1}}}=\sqrt{r_n}-\frac{\sqrt{r_{n+1}}}{\sqrt{r_{n}}}\sqrt{r_{n+1}}$$ so we need to show that $$-\frac{\sqrt{r_{n+1}}}{\sqrt{r_{n}}}\sqrt{r_{n+1}}<\sqrt{r_n}-2\sqrt{r_{n+1}}$$ which is equivalent to $$r_{n+1}>2\sqrt{r_nr_{n+1}}-r_n.$$ Using the AM-GM inequality, we have that $2\sqrt{r_nr_{n+1}}< r_{n}+r_{n+1}$. Thus $$r_{n+1}=r_{n+1}+r_n-r_n>2\sqrt{r_nr_{n+1}}-r_n$$ and the result follows.