# 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?

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?

• For d), the second: note that $\frac{a_n}{1+n^2a_n}\leq \frac{1}{n^2}$ for all $n$ and use comparison criterion. Oct 29, 2016 at 16:13
• @NikolaosSkout how do we determine if $\sum \frac{a_n}{1 + n a_n}$ converges? Certainly the trick you suggested for $\sum \frac{a_n}{1 + n^2 a_n}$ won't work here. Oct 30, 2016 at 7:02
• How do we determine if $\sum \frac{a_n}{1+n^2 a_n}$ converges? And, how do we show that $\lim_{n\to\infty} \frac{a_n}{s_n^2}$ exists? Oct 30, 2016 at 7:04
• What has caused my question to get downvoted, I wonder? Oct 31, 2016 at 17:26
• Can anybody here advise me on how to display the missing information from the comments above? Nov 28, 2016 at 5:06

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.

• how on earth did you come up with this answer? Did you read it in some book (if so, which one)? Or, did you think of it yourself? Oct 29, 2016 at 16:31
• @SaaqibMahmuud, I thought of it myself. The "difficult" case for this problem is when the $a_k$ tend to 0, and I realized that in this case $1 + a_k$ would be sufficiently close to 1 to make an argument like my second case. Then I needed to deal with when $a_k$ does not tend to 1, which gives the first case. Oct 30, 2016 at 13:29

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.

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.

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.

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.

• that's not the case, I'm afraid. For example, let's take $a_n = \frac{1}{n^2}$. Then $$\frac{a_n}{1+a_n} = \frac{1}{n^2+1}$$ and so $\sum \frac{a_n}{1+a_n}$ converges. Oct 30, 2016 at 6:48
• if $$\frac{a_n}{1+a_n} < \frac{3}{2},$$ then we have $$a_n < \frac{3}{2} + \frac{3}{2} a_n,$$ which implies $$-\frac{1}{2} a_n < \frac{3}{2}$$ and so $$a_n > \frac{\frac{3}{2}}{-\frac{1}{2}} = -3.$$ Don't you agree? Oct 30, 2016 at 6:52