Exercise 2.4.10 is about infinite products. I'd like someone to verify, if my convergence/divergence proof is technically correct and rigorous.
A close relative of the infinite series is the infinite product.
\begin{align*}
\prod_{n=1}^{\infty} b_n = b_1 b_2 b_3 \cdots
\end{align*}which is understood in terms of its sequence of partial products.
\begin{align*}
p_m = \prod_{n=1}^{m} b_n = b_1 b_2 b_3 \cdots b_m
\end{align*}Consider the special class of infinite products of the form
\begin{align*}
\prod_{n=1}^{\infty}(1+a_n) = (1+a_1)(1+a_2)(1+a_3)\cdots
\end{align*}where $a_n > 0$.
(a) Find an explicit formula for the sequence of the partial products in the case where $a_n = 1/n$ and decide whether the sequence converges. Write out the first few terms in the sequence of partial products in the case where $a_n = 1/n^2$ and make a conjecture about the convergence of this sequence.
(b) Show, in general, that the sequence of partial products converges if and only if $\sum_{n=1}^{\infty}a_n$ converges. (The inequality $1+x \le 3^x$ for positive $x$ will be useful in one direction.)
Proof.
(a) If $a_n = 1/n$, the partial product
\begin{align*}
p_m &:= \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{m}\right)\\
&= 1 + \sum_{m} \frac{1}{m} + \sum_{m < n} \frac{1}{m \cdot n} + \sum_{m < n < p} \frac{1}{m \cdot n \cdot p} + \ldots
\end{align*}
Since $\sum 1/m$ is the harmonic series, which is well-known to be divergent, this sequence of partial products is divergent.
Consider the case of $a_n = 1/n^2$. We are going to use the inequality $1 + x^2 \le e^{x^2}$. We have:
\begin{align*}
p_m &= \left(1+\frac{1}{1^2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{3^2}\right)\cdots\left(1+\frac{1}{m^2}\right)\\
&\le e^{\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots + \frac{1}{m^2}}\\
&\le e^{\frac{1}{1\cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \ldots + \frac{1}{m\cdot (m+1)}}\\
&\le e^{\left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3}-\frac{1}{4}\right) + \ldots + \left(\frac{1}{m} – \frac{1}{m+1}\right)}\\
&= e^{1 – \frac{1}{m+1}}\\
&\le e
\end{align*}
Moreover, $p_1 = 2$, $p_2 = (2)(5/4)$, $p_3 = (2)(5/4)(10/9)$, $p_4 = (2)(5/4)(10/9)(17/16)$. So, $(p_m)$ is monotonic increasing and bounded by $e$. Consequently, by the Montone Convergence Theorem, $(p_m)$ is convergent.
(b) ($\longrightarrow$) direction.
The sequence of partial products,
\begin{align*}
p_n &= (1+a_1)(1+a_2)\cdots(1+a_n)\\
&\le e^{a_1}\cdot e^{a_2}\cdots e^{a_n}\\
&\le \exp \left({\sum_{k=1}^{n}a_k}\right)
\end{align*}
If the sequence of partial products is convergent and therefore bounded, the sequence of partial sums $\sum_{k=1}^{n} a_k$ must be bounded and convergent.
Best Answer
For (a) what you have done is correct.
For (b) your argument is not valid. Note that $\sum \ln (1+a_n) <\infty$. This implies that $\ln (1+a_n) \to 0$ so $a_n \to 0$. Now, there exists $\delta >0$ such that $\ln (1+x) \geq \frac 1 2 x$ for $0<x <\delta$ (because $\lim_{x \to 0}\frac {ln (1+x)} x=1$). Hence, $a_n <2 \ln (1+a_n)$ for $n$ sufficiently large. This proves that $\sum a_n <\infty$.
The converse part follows similarly by looking at $\ln (1+a_n)$ and noting that $\ln (1+x) \leq x$ for all $x >0$.