Prob. 11, Chap. 3, in Baby Rudin: If $a_n > 0$ and $\sum a_n$ diverges, then how do we show that $\sum \frac{a_n}{1+a_n}$ too diverges?

Question

Here's Prob. 11, Chap. 3, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $a_n > 0$, $s_n = a_1 + \cdots + a_n$, and $\sum a_n$ diverges.

(a) Prove that $\sum \frac{a_n}{1+ a_n}$ diverges. [ I have no clue of how to prove this!]

(b) Prove that $$ \frac{a_{N+1}}{s_{N+1}} + \cdots + \frac{a_{N+k}}{s_{N+k}} \geq 1- \frac{s_N}{s_{N+k}}$$ [I can show this.] and deduce that $\sum \frac{a_n}{s_n}$ diverges. [How to?]

(c) Prove that $$ \frac{a_n}{s_n^2} \leq \frac{1}{s_{n-1}} - \frac{1}{s_n}$$ and deduce that $\sum \frac{a_n}{s_n^2}$ converges. [This I can show, I think.]

(d) What can be said about (the convergence or divergence of) $$\sum \frac{a_n}{1+ n a_n} \ \ \ \mbox{ and } \ \ \ \sum \frac{a_n}{1+ n^2 a_n}?$$ [ How to answer this?]

I would prefer those answers that use only the machinary developed by Rudin himself upto this point in the book.

Here's what I can show:

Since $a_n > 0$, we have $0 < s_{N+1} < \cdots < s_{N+k}$ and so $$ \frac{a_{N+1}}{s_{N+1}} + \cdots + \frac{a_{N+k}}{s_{N+k}} \geq \frac{a_{N+1} + \cdots + a_{N+k}}{s_{N+k}} = 1- \frac{s_N}{s_{N+k}}.$$ As $a_n > 0$, so, for all $n = 2, 3, 4, \ldots$, we have $0 < s_{n-1} < s_n$ and therefore $$ \frac{a_n}{s_n^2} = \frac{s_n - s_{n-1}}{s_n^2} \leq \frac{s_n - s_{n-1}}{s_n s_{n-1}} = \frac{1}{s_{n-1}} - \frac{1}{s_n},$$ and hence $$ 0 \leq \sum_{k=1}^n \frac{a_k}{s_k^2} = \frac{1}{a_1} + \sum_{k=2}^n \frac{a_k}{s_k^2} \leq \frac{1}{a_1} + \sum_{k=2}^n \left( \frac{1}{s_{k-1}} - \frac{1}{s_k} \right) = \frac{1}{a_1} + \frac{1}{s_1} - \frac{1}{s_n} \to \frac{2}{a_1} + 0 $$ as $n \to \infty$ because $a_n > 0$ and $\sum a_n$ diverges and hence $s_n = a_1 + \cdots + a_n \to \infty$ as $n \to \infty$. So if $\lim_{n\to\infty} \sum_{k=1}^n \frac{a_k}{s_k^2} $ exists [ but how to show this?}, then we must have $$ 0 \leq \lim_{n\to\infty} \sum_{k=1}^n \frac{a_k}{s_k^2} \leq \frac{2}{a_1}.$$

Am I right?

Answer 1

Consider two cases: either there are infinitely many $k$ such that $a_k \geq 2$, or there are only finitely many such $k$.

In the first case, the series $\sum_k \frac{a_k}{1+a_k}$ has infinitely many terms which are at least $\frac23$, so it diverges.

In the second case, except for finitely many $k$, we have that $$ \frac{a_k}{1+a_k} \geq \frac{a_k}3, $$ so the series also diverges.

Answer 2

Hint for $(a)$.

put $b_n=\frac{a_n}{1+a_n}$.

suppose $\sum b_n$ convergent.

then $\lim_{n\to+\infty} b_n=0$

but

$b_n=a_n(1-b_n)=a_n(1+\epsilon(n))$ thus

$a_n$ and $b_n$ are positive and equivalent when $n \to +\infty$ and $\sum a_n$ will converge which is in contradiction with the hypotheses.

Answer 3

Part (b).

For any $N$ choose $k$ such that $S_N / S_{N+k} < 1/2$. This is possible because $S_n \to \infty$.

Hence, in violation of Cauchy criterion,

$$\sum_{k=N+1}^{N+k} \frac{a_k}{S_k} \geqslant \frac{S_{N+k} - S_N}{S_{N+k}} = 1 - \frac{S_N}{S_{N+k}} > \frac{1}{2},$$

and $\sum (a_n/S_n)$ must diverge.

Answer 4

For part (a), you may consider $\alpha = lim\ sup\ a_n$. Then $a_n < 1+\alpha$ for sufficiently large $n$, so for these $n$ you have $a_n/(2+\alpha) < a_n / (1 + a_n)$.

For part (d), if you take $a_n = 1$, the series $\sum a_n/(1+na_n)$ diverges. But it can also converge. See here for an example.

Answer 5

we take $x_n$=$\frac{a_n}{1+a_n}$ . then if $\sum x_n$ is convergent then lim$x_n$=0.

so for $\epsilon$=$\frac{3}{2}$ >0 there exists k$\in$ N s.t $\forall$ n$\geqslant$k we have|$x_n$| < $\epsilon$

so |$\frac{a_n}{1+a_n}$| <$\epsilon$ so $\frac{a_n}{1+a_n}$<$\epsilon$ as $a_n$ >0 $\forall$ n$\in$ N....so

$a_n$ < $\frac{\epsilon}{1-\epsilon}$=-3 <0 $\forall$ n$\geqslant$k .but we have $a_n$ >0 $\forall$ n$\in$ N.

we have the conclution that if $a_n$ >0 then $\sum\frac{a_n}{1+a_n}$ is divergent. you need not use $\sum a_n$ is divergent.