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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?

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    $\begingroup$ 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. $\endgroup$ Commented Oct 29, 2016 at 16:13
  • $\begingroup$ @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. $\endgroup$ Commented Oct 30, 2016 at 7:02
  • $\begingroup$ 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? $\endgroup$ Commented Oct 30, 2016 at 7:04
  • $\begingroup$ What has caused my question to get downvoted, I wonder? $\endgroup$ Commented Oct 31, 2016 at 17:26
  • $\begingroup$ Can anybody here advise me on how to display the missing information from the comments above? $\endgroup$ Commented Nov 28, 2016 at 5:06

5 Answers 5

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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.

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    $\begingroup$ 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? $\endgroup$ Commented Oct 29, 2016 at 16:31
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    $\begingroup$ @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. $\endgroup$ Commented Oct 30, 2016 at 13:29
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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.

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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.

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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Oct 30, 2016 at 6:48
  • $\begingroup$ 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? $\endgroup$ Commented Oct 30, 2016 at 6:52

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