Writing a proof of the convergence of a series defined recursively

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Define the sequence $a_n$ recursively by $a_1=1$ and $$a_{n+1}=\frac13\left(a_n^2+\frac1n\right)$$

(a) Prove, by induction or otherwise, that $(a_n)$ is decreasing.
(b) Prove that the series $\sum_{n=1}^\infty (-1)^{n+1}a_n$ converges

I have been attempting this problem and have a rough answer written below. Could anyone do a solution I can use almost as a guideline as how such problems can be presented? I'm mainly asking to see how is best to layout the sort of problem as my English and formatting is a little poor.

Draft

Similar to any induction we have the base case. In this case we have P(1): We know then that $a_n=1$ and $a_{n+1}$ is $2/3$ from the formula hence $P(1)$ is true.

Then we assume $P(k)$ is true and use this to show that when $P(k)$ is true this implies $P(k+1)$ is true. We can use algebra to show that the formula holds when we have $k+1$ (i.e. it is decreasing). I do this by dragging out the inductive step and using that to show $P(k+1)$.

For the second part we have an application of the alternating series test. We have shown $a_n$ is decreasing so we need to show it converges to $0$ and is positive for all $n$ for the AST to apply then we are done.

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Base of induction

Since $a_1=1$ and $a_2=2/3$, the inequality $a_{n+1}\le a_n$ is true for $n=1$.

Inductive step

If $a_{n+1}\le a_n$, then $a_{n+1}^2\le a_n^2$, since all $a_n$ are positive by definition. Hence $$a_{n+2} = \frac13\left(a_{n+1}^2+\frac1{n+1}\right)\le \frac13\left(a_n^2+\frac1n\right) = a_{n+1}$$ which establishes the inductive step.

Convergence of the series

In order to apply the Alternating Series Test we need $a_n>0$ (done), $a_{n}$ decreasing (done), and $a_n\to 0$. It remains to show the last property. Since the sequence $(a_n)$ is decreasing and bounded below, it has a limit, call it $L$. Then $$ \lim_{n\to \infty}\frac13\left(a_n^2+\frac1n\right) = \frac13 L^2 $$ On the other hand, this is just $\lim_{n\to \infty}a_{n+1}$, the limit of the same sequence with index shifted by one. So, $L=\frac13 L^2$. This means either $L=0$ or $L=3$. The latter is impossible because $a_1=1$ and the sequence is decreasing.