# Writing a proof of the convergence of a series defined recursively

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.

• Can you tell us what your rough answer is? Dec 18, 2014 at 17:16
• Sure, 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.. Dec 18, 2014 at 19:22
• ..for the AST to apply then we are done. I'm mainly asking to see how is best to layout the sort of problem as my english and formatting is a little poor. I'm not trying to get someone else to do my work if that is what you are thinking I just want to get my presentation the best it can be!!!! Dec 18, 2014 at 19:28

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