# Convergence of sum given convergence of sequence

Let $$\{a_n\}_{n\in\mathbb{N}}$$ the sequence defined by $$a_1 = \tfrac{1}{2}$$ and $$a_{n+1}=\sqrt{\tfrac{1}{2}+a_n}$$ for $$n\geq 1$$. Prove that the sequence converges and, if $$L$$ is the limit of the sequence, prove that $$\sum_{n=1}^\infty |a_n - L|$$ converges.

I have proven that the limit $$L$$ of the sequence $$\{a_n\}_{n\in\mathbb{N}}$$ exists checking that is an increasing sequence and that is bounded (in particular, $$a_n\in [0,2]$$ for any $$n\geq 1$$). Then, for every $$\varepsilon >0$$, there exists $$n_0\in\mathbb{N}$$ such that for any $$n\geq n_0$$ then $$|a_n - L|<\varepsilon$$.

Using this in the sum, for every $$\varepsilon >0$$, there exists $$n_0\in\mathbb{N}$$ such that $$\sum_{n=1}^\infty |a_n - L| = \sum_{n=1}^{n_0} |a_n - L| + \sum_{n=n_0+ 1}^\infty |a_n - L| < \sum_{n=1}^{n_0} |a_n - L| + \sum_{n=n_0+ 1}^\infty \varepsilon$$

However, I don't know how to finish the proof. Also, I would be interested if the convergence of the sum is true for any convergent sequence $$\{a_n\}_{n\in\mathbb{N}}$$.

Any help would be appreciated.

We know $$L^2=\frac 12+L, L=\frac{1+\sqrt 3}{2}>1$$ and $$a_n > 0$$. Then $$a_{n+1}^2-L^2=\frac 12+a_n-L^2=\frac 12 + a_n - \left(\frac 12 +L\right) = a_n-L\\ \implies |a_{n+1}-L|=\frac{|a_n-L|}{a_{n+1}+L} < \frac{|a_n-L|}{L}.$$