# Prove or disprove the recursively defined sequence is convergent.

The sequence $$\{a_n\}$$ is defined by $$a_1=1, a_2=0$$ and $$a_{n+2}=a_{n+1}+\displaystyle\frac{a_n}{n^2}$$ for $$n\in \mathbb{N}$$.

Since $$\displaystyle\frac{1}{n^2}$$ is summable, when $$n$$ is large, the sequence is something like $$a_n=a_{n-1}+\displaystyle\sum_{i\leq n-2}\frac{a_i}{i^2}$$, so I think the sequence should be convergent.

Then I want to use the Monotone convergent theorem, i.e. to show $$\{a_n\}$$ is monotonic and bounded.

For monotonic, it is easy to see that $$\{a_n\}$$ is increasing.

But for the upper bound, assuming $$\{a_n\}$$ converges and taking the limit $$n\to \infty$$ does not give any hints for me to find a suitable upper bound. I have also used computer programs to compute up to the 10000th term, but it seems that $$\{a_n\}$$ is still increasing, does not converges to a certain number.

So I wonder if it is convergent or not.

Well this took longer than I thought. I feel like there must be an easier solution...

Claim 1: $$a_n\le\sqrt n$$ for all $$n$$. This holds for $$n=1$$ and $$n=2$$. Actually, we will want to assume $$n\ge 3$$ later, so we can also check $$a_3=1\le\sqrt3$$. Now if $$a_n\le\sqrt n$$ and $$a_{n+1}\le\sqrt{n+1}$$, then

$$a_{n+2}=a_{n+1}+\frac{a_n}{n^2}\le\sqrt{n+1}+\frac{\sqrt n}{n^2},$$

and it suffices to show $$\sqrt{n+1}+\frac{\sqrt n}{n^2}\le \sqrt{n+2}$$. Note

$$\sqrt{n+1}+\frac{\sqrt n}{n^2}\le\sqrt{n+1}+\frac{\sqrt{n+1}}{n^2}=\sqrt{n+1}\left(1+\frac{1}{n^2}\right)$$

and the inequality

$$\sqrt{n+1}\left(1+\frac{1}{n^2}\right)\le\sqrt{n+2}$$

is equivalent to

$$(n+1)\left(1+\frac{1}{n^2}\right)^2\le n+2.$$

With some elbow grease this is equivalent to

$$n^4\ge 2n^3+2n^2+n+1.$$

Now since $$n\ge 3$$,

$$n^4\ge 3n^3=2n^3+n^3\ge 2n^3+3n^2=2n^3+2n^2+n^2\ge 2n^3+2n^2+n+1.$$

This establishes Claim 1.

Claim 2: $$a_n=\sum_{i=1}^{n-2}\frac{a_i}{i^2}$$ for $$n\ge 3$$. This holds for $$n=3$$, and if $$a_n=\sum_{i=1}^{n-2}\frac{a_i}{i^2}$$, then $$a_{n+1}=a_n+\frac{a_{n-1}}{(n-1)^2}=\frac{a_{n-1}}{(n-1)^2}+\sum_{i=1}^{n-2}\frac{a_i}{i^2}=\sum_{i=1}^{n-1}\frac{a_i}{i^2}.$$

Finishing up: we now have $$a_n=\sum_{i=1}^{n-2}\frac{a_i}{i^2}\le\sum_{i=1}^{n-2}n^{-\frac32}.$$ Pick your favorite way to show this is the partial sum of a convergent $$p$$-series, and we're done!

Hint: Show that $$a_n \leq \prod_{i \leq n} (1+\frac 1{i^{3/2}})$$ for all $$n \geq 4$$. The infinite product $$\prod_{i \leq n} (1+\frac 1{i^{3/2}})$$ is convergent because $$\sum_n \frac 1 {n^{3/2}} <\infty$$.

[ Above inequality may hold for $$n <4$$ also but I found it easy to verify it for $$n \geq 4$$].

• @ElliotG Thanks for pointing out. – Kavi Rama Murthy Oct 1 at 8:11
• @ashim03217 I have found a proof for boundedness and I have edited my answer. – Kavi Rama Murthy Oct 1 at 8:47
• Okay I will try it, thanks for your help. Btw, I wonder how you come up with such bound? @Kavi Rama Murthy – ashim0317 Oct 1 at 9:02
• I was trying to get a bound $a_n \leq c_n$ with $c_n$ convergent. If $a_n$ and $a_{n+1}$ are both $\leq d_n$ then $a_{n+2} \leq d_n (1+\frac 1 {n^{2}})$. This led to consideration of $\prod (1+\frac 1 {n^{2}})$ but that didn't wort. So I lowered the power $n^{2}$ to $n^{1.5}$. @ashim0317 – Kavi Rama Murthy Oct 1 at 9:07

Here is yet another approach: Clearly $$a_n\ge 0$$ for all $$n$$, so we can derive the following for all $$n>1$$: $$a_{n}\le a_{n+1}$$ and therefore $$a_{n+2} \le a_{n+1} + a_{n+1}/n^2$$, or equivalently $$\frac{a_{n+2}}{a_{n+1}}-1 \le \frac{1}{n^2}.$$ For all positive $$x$$, we have $$\log x\le x-1$$ and so it follows that $$\log\frac{a_{n+2}}{a_{n+1}} \le \frac{1}{n^2}.$$ Since all the terms $$\log(a_{n+2}/a_{n+1})$$ are non-negative and are dominated by the sequence $$1/n^2$$ whose sum converges, the sum of $$\log(a_{n+2}/a_{n+1})$$ converges too. But its partial sum from $$n=2$$ to $$n=m$$ is simply $$\log(a_{m+2}/a_{3})$$, so $$a_{m+2}$$ clearly converges.