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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.

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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!

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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 $n \geq 4$].

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  • $\begingroup$ @ElliotG Thanks for pointing out. $\endgroup$ – Kavi Rama Murthy Oct 1 at 8:11
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    $\begingroup$ @ashim03217 I have found a proof for boundedness and I have edited my answer. $\endgroup$ – Kavi Rama Murthy Oct 1 at 8:47
  • $\begingroup$ Okay I will try it, thanks for your help. Btw, I wonder how you come up with such bound? @Kavi Rama Murthy $\endgroup$ – ashim0317 Oct 1 at 9:02
  • $\begingroup$ 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 $\endgroup$ – Kavi Rama Murthy Oct 1 at 9:07
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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.

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