# 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?

The first part is correct, but you have left out a lot of details. For example, why is rn=s-sn-1? You cannot simply add and subtract, it's a limiting process. For n>m, m fixed, sn-sm-1 tends to rm. This limiting value is also equal to s-sm-1.

For the second part, use the comparison test.

• 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? Commented Oct 30, 2016 at 9:57
• If you go through Rudin's chapter on sequences and series, you will see that he defines the limit of a series as the limit of a sequence.
– user384095
Commented Oct 30, 2016 at 13:47
• Suppose s<sub>n</sub> tends to s. Then s<sub>n</sub>-s<sub>m-1</sub> tends to r<sub>m</sub>, where the initial value of n>m. But since m is fixed, the limiting value is also equal to s-s<sub>m</sub>.
– user384095
Commented Oct 30, 2016 at 13:57
• 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. Commented Oct 30, 2016 at 14:04
• Yes, that is what I meant. The point is that you have to explicitly prove what you are claiming, especially with a series. Also, the only reason the equality holds is because the terms are all positive. If you already knew the proof to this, then my apologies.
– user384095
Commented Oct 30, 2016 at 14:13