Exercise 3.12

Question

Exercise 12: Suppose $a_n>0$ and $\sum a_n$ converges. Put $$r_n=\sum_{m=n}^\infty a_n.$$ (a) Prove that $$\frac{a_m}{r_m}+\ldots+\frac{a_n}{r_n}>1-\frac{r_n}{r_m}$$ if $m

ghostofgarborg · Accepted Answer

(a) Since $r_n$ is monotonically decreasing $$ \frac{a_m}{r_m} + \cdots + \frac{a_n}{r_n} > \frac{a_{m}}{r_m} + \cdots + \frac{a_{n-1}}{r_m} = \frac{a_{m} + \cdots + a_{n-1}}{r_m} = \frac{r_{m} - r_n}{r_m} = 1 - \frac{r_n}{r_m} $$ Since $r_n \to 0$, given any $M$ we can find an $N > M$ such that $1 - \frac {R_N}{R_M} $ is arbitrarily close to $1$. This implies that $a_n/r_n$ is not Cauchy, hence not convergent. (b) $$ \frac{a_n}{\sqrt{r_n}} = \frac{a_n \left( 1 + \frac{\sqrt{r_{n+1}}}{\sqrt{r_n}} \right)}{\sqrt{r_{n}} + \sqrt{r_{n+1}}} < \frac{2a_n}{\sqrt{r_{n}} + \sqrt{r_{n+1}}} = \frac{2(r_{n} - r_{n+1})}{\sqrt{r_{n}} + \sqrt{r_{n+1}}} = 2(\sqrt{r_{n}} - \sqrt{r_{n+1}}) $$ Since $$ \sum_{n=1}^N 2(\sqrt{r_{n}} - \sqrt{r_{n+1}}) = 2(\sqrt {r_{1}} - \sqrt{r_{N+1}}) < 2\sqrt{r_1}$$ the series converges by the comparison test.

Exercise 3.12

Answers

Comments

Exercise 3.12

Answers

Comments

Add answer