Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.