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 $$ a_{n+1}^2-L^2=\frac 12+a_n-L^2=\frac 12 + a_n - \left(\frac 12 +L\right) = a_n-L\\ \implies |a_{n+1}-L|=\frac{|a_n-L|}{a_{n+1}+L} < \frac{|a_n-L|}{L}. $$

Can you end it now?

  • $\begingroup$ Brilliant answer. Thank you! $\endgroup$
    – user326159
    Dec 6 '20 at 17:47

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