Limit of $n^2\sqrt{1-\cos(1/n)+\sqrt{1-\cos(1/n)+\ldots}}$ when $n \to \infty$ Compute the limit:
$$\lim_{n \to \infty} n^2\sqrt{1-\cos(1/n)+\sqrt{1-\cos(1/n)+\sqrt{1-\cos(1/n)+\ldots}}}$$
 A: Hints:


*

*For every $a\gt0$, $b=\sqrt{a+\sqrt{a+\sqrt{a+\cdots}}}$ is such that $b^2+a=b$ and $b\gt0$, thus $b=\frac12+\frac12\sqrt{1+4a}$.

*When $n\to\infty$, $1-\cos(1/n)\to0$.

*When $a\to0$, $\frac12+\frac12\sqrt{1+4a}\to1$.

*Hence the limit you are after is $\lim\limits_{n\to\infty}n^2\cdot1=+\infty$.

A: Define
$$
a_m=\underbrace{\small\sqrt{1-\cos(1/n)+\sqrt{1-\cos(1/n)+\sqrt{1-\cos(1/n)+\ldots+\sqrt{1-\cos(1/n)}}}}}_{m\text{ square roots}}
$$
$a_1=\sqrt{1-\cos(1/n)}=\sqrt{2\sin^2\left(\frac1{2n}\right)}\lt2$. If $a_m\le2$, then
$$
\begin{align}
a_{m+1}
&=\sqrt{1-\cos(1/n)+a_m}\\
&\le\sqrt{2+2}\\[4pt]
&=2
\end{align}
$$
Furthermore, note that $a_{m+1}\ge a_m$. Thus, $a_m$ is an increasing sequence, bounded above; therefore,
$$
A_n=\lim_{m\to\infty}a_m
$$
exists, is non-negative, and is no greater than $2$. We then have
$$
A_n^2=2\sin^2\left(\frac1{2n}\right)+A_n
$$
Solving for $A_n$ yields
$$
A_n=\frac{1+\sqrt{1+8\sin^2\left(\frac1{2n}\right)}}{2}
$$
$A_n\ge1$ for all $n$, therefore,
$$
\lim_{n\to\infty}n^2A_n=\infty
$$
