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.

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.
