3
$\begingroup$

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.

$\endgroup$
3
  • 2
    $\begingroup$ Can you tell us what your rough answer is? $\endgroup$
    – Clement C.
    Dec 18, 2014 at 17:16
  • $\begingroup$ 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.. $\endgroup$
    – Alex R
    Dec 18, 2014 at 19:22
  • $\begingroup$ ..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!!!! $\endgroup$
    – Alex R
    Dec 18, 2014 at 19:28

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.