<s>[INST] solution verification here is solution of my old question but i can't see it would someone explain to me the principal idea and what he wants to show
Solution
from $u_n=\sqrt{n}\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right),$ To $u^2_n=\frac{n}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right),$
and from
$u^2_n= \frac{n}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right)$
To $u^2_n=\frac{n}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)$
and if we've that $u_n^2$ convergent is that means $u_n$ also convergent
i just need to see the principal idea  of what he did Thanks
I think the principal idea of solution is that he want to show $u_n^2$ convergent and bounded then tell that $u_n$ is convergent too
Any help would be appreciated
 [/INST] 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 $u_n>0$.
</s>