Prob. 12, Chap. 3 in Baby Rudin: Some results involving the remainder of a convergent series of positive term series

Here is Prob. 12, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

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

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

Prove that $$\frac{a_n}{\sqrt{r_n}} < 2 \left( \sqrt{r_n} - \sqrt{r_{n+1}} \right)$$ and deduce that $\sum \frac{a_n}{\sqrt{r_n}}$ converges.

My effort:

Let $s_n = a_1 + \cdots + a_n$, and let $s = \sum_{n=1}^\infty a_n$. Then $r_n = s - s_{n-1}$.

Moreover, since $s_n \to s$ as $n \to \infty$, so $r_n \to 0$.

Also as $a_n > 0$, we can conclude that $s_n < s_{n+1}$ and also that $s = \sup \left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\}$, which implies that $r_n > 0$. [Am I right?]

If $m < n$, then $s_{m-1} < s_{n-1}$ and so $r_m > r_n > 0$. So $$\frac{a_m}{r_m} + \cdots + \frac{a_n}{r_n} > \frac{ a_m + \cdots + a_n}{r_m} = \frac{ s_n - s_{m-1}}{r_m} = \frac{ r_m - r_{n+1} }{r_m} = 1 - \frac{ r_{n+1}}{r_m} > 1- \frac{r_n}{r_m}.$$

Fix $m \in N$. Now as $\lim_{n \to \infty} r_n = 0$, so we can find a natural number $N$ such that $n > N$ implies $r_n < \frac{r_m}{2}$. So, for any natural number $M$, we can find natural numbers $n$, $m$, such that $n > m > M$, but $$\sum_{k=m}^n \frac{a_k}{r_k} > \frac{1}{2}.$$ So the Cauchy criterion for convergence is not satisfied by the series $\sum \frac{a_n}{r_n}$, which therefore diverges. Am I right?

Now we note that $$2 \left( \sqrt{ r_n} - \sqrt{r_{n+1}} \right) = \frac{ 2 \left( r_n - r_{n+1} \right)}{ \sqrt{ r_n} + \sqrt{r_{n+1}} } > \frac{ 2 \left( s- s_{n-1} - s + s_n \right)}{ 2 \sqrt{r_n}} = \frac{a_n}{\sqrt{r_n}}.$$ Am I right?

Thus, for all $n \in \mathbb{N}$, we have $$0 \leq \sum_{k=1}^n \frac{a_k}{\sqrt{r_k}} < 2 \sum_{k=1}^n \left( \sqrt{r_k} - \sqrt{r_{k+1}} \right) = 2 \left( \sqrt{r_1} - \sqrt{r_{n+1}} \right) = 2 \left( \sqrt{s} - \sqrt{ s - s_n} \right) \to 2 \sqrt{s}$$ as $n \to \infty$ and therefore $$0 \leq \lim_{n\to \infty} \sum_{k=1}^n \frac{a_n}{\sqrt{r_n}} \leq 2 \sqrt{s},$$ provided that this limit exists. But how do we show that this limit does exist?

• how then would we justify $r_n = s-s_{n-1}$? Or does this relation hold at all? If it doesn't, then the whole of my argument collapses. Can you please supply the details in your answer? – Saaqib Mahmood Oct 30 '16 at 9:57
• I'll try to write it out for you. Suppose $s_n \to s$. Then $s_n - s_{m-1} \to r_m$, where the initial value of $n > m$. But since $m$ is fixed, the limiting value is also equal to $s-s_m$. Is this waht you wanted to write? If so, I didn't get your point. – Saaqib Mahmood Oct 30 '16 at 14:04