Here's the question, and I want to know if my steps are correct:
Assume $x_i \geq 0$. If $\sum_i^\infty x_i$ converges, show that $\prod_i^\infty(1+x_i)$ converges as well.
Attempt: Since $\prod_i^\infty(1+x_i)$ is monotone increasing, we just need to find an upper bound for $\prod_i^\infty(1+x_i)$.
Suppose $\sum_i^\infty x_i=L$, then we have $\sum_i^M x_i \leq L$ for some $M \in \mathbb{N}$. Using AM-GM, we obtain:
\begin{align} \left( \prod_i^M(1+x_i)\right)^\frac{1}{M} &\leq \frac{(1+x_1)+(1+x_2)+\cdots+(1+x_M)}{M} \\ & \leq \frac{M+\sum_i^M x_i}{M} \\ &\leq \frac{M+L}{M} \\ &= 1+\frac{L}{M} \end{align}
Raising both sides by the power $M$ and taking limits of both sides,
\begin{align} \prod_i^M(1+x_i) &\leq \left(1+\frac{L}{M}\right)^M \\ \lim_{M \rightarrow \infty} \prod_i^M(1+x_i) & \leq \lim_{M \rightarrow \infty}\left(1+\frac{L}{M}\right)^M \\ \lim_{M \rightarrow \infty} \prod_i^M(1+x_i) & \leq e^L \end{align}
Is my proof correct? I'm worried about the step which I take the limits of both sides of the inequality.
You write $\prod^\infty_i(1+x_i)$ is monotone increasing.
But $\prod^\infty_i(1+x_i)$ is merely a number. What does it mean to say a number is monotone increasing? What I presume you mean is $$ \prod_{i=1}^M (1+x_i) $$ (with $M$ rather than $\infty$ in the superscript) is a monotone increasing function of $M$.
You say $\sum^M_{i=1} x_i≤L$ for some $M\in\mathbb N.$ But here you ought to say for all $M\in \mathbb N.$
Other than that your argument is good.
I also like this way: $$ \prod_{i=1}^M (1+x_i) \le \prod_{i=1}^M e^{x_i} = e^{\sum_{i=1}^M x_i} $$ and so on.