Showing convergence of a series Let $a_0,a_1\in\mathbb{R}$,$a_n=a_{n-1}-\cfrac{2}{n}a_{n-2}$ for $n\ge2$.
How to show $\sum_0^\infty|a_n|<\infty$
 A: Use generating functions.  Set $f(x) = \sum_{n=0}^\infty a_n x^n$.  Then plugging this into the recurrence relation, you will end up with
$$ (1-x)f'(x) + (2x-1)f(x) = (a_1-a_0) .$$
Solve this using integrating factors, and you will get
$$ f(x) = (1-x)e^{2x}\left(C_1 + (a_1-a_0)\int \frac{e^{-2x} \, dx}{(1-x)^2} \right) .$$
Now $(1-x)e^{2x}$ is an entire function, and hence has infinite radius of convergence, and hence the coefficients are easily seen to be absolutely summing.
Now for $(1-x)e^{2x}\int \frac{e^{-2x} \, dx}{(1-x)^2}$, integrate by parts twice.  One term will look like $(1-x)\ln(1-x)$, and this has coefficients like $1/(n(n+1))$, which is absolutely summable.  The other term will look like $(1-x)e^{2x}\int e^{-2x} \ln(1-x) \, dx$.  The integrand has terms that decay like $1/n$, and hence after integrating they will decay like $1/n^2$.  Multiplication by $e^{2x}$ or $e^{-2x}$ will have negligible effect on the decay of coefficients.
I think this is a really hard problem (unless there is some other, easy way, which I did not see).  Where did this problem come from?
Here is some Mathematica code that illustrates that the generating function is correct.

A: Prologue - Since $\frac{a_{n} - a_{n-1}}{n+1}$ is somewhat "small", it is reasonable to think that $(a_n)$ will roughly behave like the sequence $(b_n)$  with $b_2=a_2$ satisfying for all $n\geq 2$,
$$
b_{n+1} = b_{n}-\frac{2}{n+1}b_{n}=\left(1-\frac{2}{n+1}\right)b_{n}
$$
for which we have
$$
|b_n| \leq |b_2|\prod_{k=3}^n\left(1-\frac{2}{k}\right)\leq |b_2|\exp\left(-2\sum_{k=3}^n \frac{1}{k}\right) = O\left(\frac{1}{n^2}\right).$$
We can already prove the very crude bound $a_n = O(n^2)$ using these basic ideas. Let $c_0=|a_0|$ and $c_1=|a_1|$ and for $n \geq 2$, $c_{n} = c_{n-1} + \frac{2}{n}c_{n-2}$. It is clear that $|a_n|$ is bounded by $c_n$. Also $c_n$ is increasing, so that $c_{n+1} \leq (1+2/(n+1))c_n$, leading to $c_n = O(n^2)$.
Act I - Let us prove that $a_n = O(1)$.
Since we showed $a_n = O(n^2)$, the "error" sequence $\epsilon_n = \dfrac{4a_{n-2}}{n(n+1)}$ is bounded, and
$$
\tag{$\ast$}
\forall n \geq 2,\qquad a_{n+1} = \left(1-\frac{2}{n+1}\right)a_n + \epsilon_n.
$$
Considering the inequality $|a_{n+1}| \leq |a_n| + |\epsilon_n|$, we deduce from $\epsilon_n=O(1)$ that $a_n = O(n)$, which in turn implies $\epsilon_n = O(1/n)$, hence $a_n = O(\log n)$ and $\epsilon_n = O((\log n)/n^2)$.
Finally, $a_n = O(1)$ because $\sum|\epsilon_n| < \infty$.
Of course it is a good starting point, but we can do much better.
Act II - Actually, $a_n = O\left(\dfrac{1}{n^2}\right)$.
Using ($\ast$) again, the inequality
$$
\sum_{n=2}^N\frac{2|a_n|}{n+1} \leq \sum_{n=2}^N(|a_{n}|-|a_{n+1}|+ |\epsilon_n|) \leq |a_N|+\sum_{n=2}^\infty |\epsilon_n| = O(1)
$$
shows that $\sum \frac{|a_n|}{n} < \infty$, and $a_N = a_2 + \sum_{n=2}^{N-1}\frac{-2a_n}{n}$ is convergent as $N \to \infty$. The condition $\sum \frac{|a_n|}{n} < \infty$ shows that $\lim a_n = 0$.
From ($\ast$), we can write $(n+1)|a_{n+1}| - n|a_n| \leq\frac{4|a_n|}{n}$ and sum for $2\leq n\leq N-1$ in order to prove that $n|a_n| = O(1)$.
From ($\ast$), we can also write $(n+1)^2|a_{n+1}| \leq n^2|a_n|+7|a_n|$ and sum for $2\leq n\leq N-1$:
$$
N^2|a_N| - 4|a_2| \leq 7 \sum_{n=2}^{N-1} |a_n| = O\left(\sum_{n=2}^{N-1} \frac{1}{n}\right) = O(\log N),
$$
and this estimate automatically improves to $N^2|a_N|-4|a_2| \leq O\left(\sum_n \frac{\log n}{n^2}\right) =O(1)$.
Act III - In the end, we have proved that there exists some constant $C > 0$ such that $a_n \leq \dfrac{C}{n^2}$, so that the conclusion writes
$$
\sum_{n\geq 2} |a_n| \leq \sum_{n\geq 2} \frac{C}{n^2} < \infty.
$$
A: The generating function is great, it gives the general approach for this of problem.

Here is my proof for this problem only.
.
We'll prove these below statement consicutively:
i) $\{a_n \}$ converges $0$ 
ii) $|a_n| \le \frac{1}{n}$ for sufficiently large $n$.
iii) $ |a_n| \le 7\frac{ln(n) }{n^2}$ for sufficiently large $n$.
iv) $ \lim_{n=1}^{\infty} \frac{ln(n) }{n^2}$ converges
Thus $ \lim_{n \rightarrow \infty} |a_n|$ converges.

Proof for i):
Unfortunately, i only have a rather long proof for this, but I think other would have a much simple one, thus I leave it ommited.
Proof for ii):
We have: $ 2a_{n-2}-a_{n-1}= (n-1)a_{n-1}-n.a_n$
$\Leftrightarrow  2na_{n-2}-n.a_{n-1}=(n-1)na_{n-1}-n^2.a_n$
$\Leftrightarrow  2na_{n-2}-na_{n-1}-(n-1)a_{n-1}= (n-1)^2.a_{n-1}-n^2.a_n$
$\Leftrightarrow  2n( a_{n-2}-a_{n-1})+a_{n-1}=(n-1)^2.a_{n-1}-n^2.a_n$
$\Leftrightarrow  \frac{4n}{n-1}(-a_{n-3})+a_{n-1}=(n-1)^2.a_{n-1}-n^2.a_n$

Thus , $\lim_{n \rightarrow \infty}(n-1)^2a_{n-1}-n^2.a_n = 0$.
Therefore ,$\exists N >0:  \frac{1}{2} > |(n-1)^2a_{n-1}-n^2.a_n| \forall n >N$.
 Hence $\frac{1}{2}.(m-n) > | m^2.a_m-n^2.a_n| \forall m,n >N$
 $\Rightarrow \frac{1}{2}(1-\frac{n}{m}) > | m.a_m-\frac{n^2}{m}.a_n| \forall m,n>N$
 Let $m \rightarrow \infty$.We got the desire ineq.

Proof for iii)

Once again, we have :
$ \frac{4n}{n-1}(-a_{n-3})+a_{n-1}=(n-1)^2.a_{n-1}-n^2.a_n$
$\Leftrightarrow \frac{1}{n.ln(1+\frac{1}{n-1})}.n|\frac{4n}{n-1}(-a_{n-3})+a_{n-1}|=\frac{|(n-1)^2.a_{n-1}-n^2.a_n|}{ln(n)-ln(n-1)} $

Due to 2 and the fact that $\lim n.ln(1+\frac{1}{n})=1$,we imply :
 $\exists N_1>0: \forall n>N_1 \frac{|(n-1)^2.a_{n-1}-n^2.a_n|}{ln(n)-ln(n-1)} <6$
$ \Rightarrow |m^2.a_m -n^2.a_n|<6.|ln(m)-ln(n)| \forall m,n> N_1$
$\Rightarrow \exists N_2>0: |a_n| < 7.\frac{\ln(n)}{n^2} \forall n>N_2$
Proof for iv)
It's nearly obvious.
So forth.
QED.
