$\{u_n^2\}$ is increasing

Question

how he does Know that $\{u_n^2\}$ is increasing this question is related to that one solution verification

here is his solution

Note first that $$ 1-\frac{1}{k}+\frac{1}{4k^2} =\left(1-\frac{1}{k}\right)\left(1+\frac{k}{k-1}\cdot\frac{1}{4k^2}\right)= \left(1-\frac{1}{k}\right)\left(1+\frac{1}{4(k-1)k}\right). $$ Next, $$ \frac{n}{4}\prod_{k=2}^n\left(1-\frac{1}{k}\right)=\frac{n}{4}\prod_{k=2}^n \frac{k-1}{k}=\frac{1}{4}. $$ Then $$ \prod_{k=2}^n\left(1+\frac{1}{4k(k-1)}\right)=\exp\left(\sum_{k=2}^n \log\left(1+\frac{1}{4k(k-1)}\right)\right)\le\exp\left(\sum_{k=2}^n \frac{1}{4k(k-1)}\right), $$ and $$ \sum_{k=2}^n \frac{1}{k(k-1)}=\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=1-\frac{1}{n}<1. $$ Thus $$ \prod_{k=2}^n\left(1+\frac{1}{4k(k-1)}\right)\le \exp\left(\frac{1}{4}\right) $$ and hence $\{u_n^2\}$ is upper bounded and increasing, and thus convergent, and so is $\{u_n\}$, as $u_n>0$.

Answer 1

$$b_n=\prod_{k=1}^{n} (1+a_k)>1$$

Assume each $a_k$ to be positive.

$$b_{n+1}=\prod_{k=1}^{n+1} (1+a_k)=b_n(1+a_{n+1})>b_n$$