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Let $a_1=\sqrt{2}$, $a_2=\sqrt{2+\sqrt{2}}$ and $a_n$ be defined as $$a_n=\underbrace{\sqrt{2+\sqrt{2+...+\sqrt{2}}}}_{n-times}$$ for any $n\geq1$. Now consider the following infinite sum: $$\sum^{\infty}_{n=0}(a_n-2)$$ Does this sum converge? If so what would be its limit? It is easy to show that $\lim_{n\to\infty}(a_n-2)=0$ but this does not ensure us convergence of the sum. In general I would like to address the following question. Given a nested radical of the following form $$a_n=\underbrace{\sqrt[k]{b+\sqrt[k]{b+...+\sqrt[k]{b}}}}_{n-times}$$ where $k\in\mathbb{N}$ and $b\in\mathbb{R^+}$ such that $$\lim_{n\to\infty}a_n=L$$ What could we say about the convergence of the infinite sum below $$\sum^{\infty}_{n=1}(a_n-L)$$ If it converges does it posses a closed form? Is there a general result on these series?

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Let $f(x)=\sqrt{2+x}$. Then $a_{n+1}=f(a_n)$, $a_1=\sqrt2$. $f(2)=2$ and $f'(2)=1/4$. There exists $n_0$ such that $$ n\ge n_0\implies |f'(a_n)|\le\frac12. $$ Then, if $n>n_0$, by the mean value theorem we have $$ |a_n-2|=|f(a_{n-1)}-f(2)|\le\frac12\,|a_{n-1}-2|. $$ Iterating we get $$ |a_n-2|\le2^{n_0-n}|a_{n_0}-2|. $$ The series $\sum(a_n-2)$ is thus convergent.

The argument can be applied to any sequence defined by a recurrence $a_{n+1}=f(a_n)$ suche that $f(\ell)=\ell$, $a_n\to\ell$ and $|f'(\ell)|<1$.

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