Exercise 3.11

Question

Exercise 11:
    Suppose $a_n>0$, $s_n=a_1+\cdots+a_n$, and $\sum a_n$ diverges.

(a) Prove that $\sum\bigl(a_n/(1+a_n)\bigr)$ diverges.

(b) Prove that $$\frac{a_{N+1}}{s_{N+1}}+\cdots+\frac{a_{N+k}}{s_{N+k}}\ge 1-\frac{s_N}{s_{N+k}}$$ and deduce that $\sum(a_n/s_n)$ diverges.

(c) Prove that $$\frac{a_n}{s_n^2}\le\frac{1}{s_{n-1}}-\frac{1}{s_n}$$ and deduce that $\sum(a_n/s_n^2)$ converges.

(d) What can be said about $$\sum\frac{a_n}{1+na_n}\quad\hbox{and}\quad\sum\frac{a_n}{1+n^2a_n}?$$

ghostofgarborg · Accepted Answer

(a) If $a_n \leq 1$, then $\frac{a_n}{1+a_n} \geq \frac{a_n}{2}$.
    If $a_n > 1$, then $\frac{a_n}{1+a_n} > \frac{1}{2}$. Therefore
    $$ \sum \frac{a_n}{1+a_n} \geq \frac 1 2 \sum \min(a_n, 1) $$
    If $ \{ n : a_n > 1 \} $ is infinite, this clearly diverges. Otherwise, there is an 
    $N$ such that $ n > N $ implies $a_n \leq 1$, in which case the series diverges by
    comparison to $\frac 1 2 \sum a_n$.

(b) Since $S_n$ is monotonically increasing, whenever $ j \leq k $
    $$ \frac{a_{n+j}}{S_{n+j}} \geq \frac{a_{n+j}}{S_{n+k}} $$
    Consequently 
    $$ \frac{a_{N+1}}{S_{N+1}} + \cdots + \frac{a_{N+k}}{S_{N+k}} 
        \geq \frac{a_{N+1}}{S_{N+k}} + \cdots + \frac{a+{N+k}}{S_{N+k}} 
        = \frac{a_{N+1} + \cdots + a_{N+k}}{S_{N+k}} 
        = 1 - \frac{S_{N}}{S_{N+k}}$$
    Since $S_n$ is increasing and unbounded, it is possible to choose $k$ such that 
    $S_{N}/S_{N+k}$ is arbitrarily close to $0$. This shows that $\frac{a_n}{S_n}$ 
    is not Cauchy, and therefore not convergent.

(c) Since $S_n$ is monotonically increasing, we have that $ S_n^2 \geq S_{n-1}S_n $, so we can deduce that
    $$ \frac{a_n}{S_n^2} \leq \frac{S_n - S_{n-1}}{S_{n-1}S_n} = \frac{1}{S_{n-1}} - \frac{1}{S_{n}} $$
    Since $1/S_n$ is bounded 
    $$ \sum_{n=1}^N \frac{a_n}{S_n^2} \leq \frac{a_1}{S_1^2} + \sum_{n=2}^N \left( \frac 1 {S_{n-1}} - \frac 1 {S_n} \right) \leq \frac{a_1}{S_1^2} + \frac 1 {S_{1}} - \frac 1 {S_N} $$
    so $\sum a_n/S_n^2$ is monotonic and bounded, hence convergent.

(d) Since 
    $$ \frac{a_n}{1 + n^2a_n} \leq \frac{1}{n^2} $$
    $\sum a_n/(1+n^2a_n)$ converges. The series $\sum a_n/(1+na_n)$, however, might converge or diverge. 
    Let $a_n = \frac 1 n$, and it is clear that it diverges. Let $a_n = 1$ whenever $n$ is a square
    and $a_n = 2^{-n}$ otherwise. This series clearly diverges, since the terms do not tend to 0 
   as $n \to \infty$. Then 
    $$ \sum_{n=1}^N \frac{a_n}{1+na_n} \leq \sum_{n=1}^N \frac{1}{n^2} + \sum_{n=1}^{N} \frac 1 {2^n} $$
    and the series therefore converges.

Exercise 3.11

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Exercise 3.11

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