show this sum is $\sum_{n=0}^{+\infty}|a_{n}|<+\infty$ define sequence $a_{n}$  if $a_{0},a_{1}$ be arbitrary real number ,and such $$a_{n}=a_{n-1}-\dfrac{2}{n}a_{n-2}$$
show that
$$\sum_{n=0}^{+\infty}|a_{n}|<+\infty$$
I remember seeing someone asking this question before.But I can't find it,can you help or solve this problem?Thanks
Try:
$$a^2_{n}-a^2_{n-1}=-\dfrac{2}{n}a_{n-2}(a_{n}+a_{n-1})$$
so we have
$$\sum_{i=1}^{n}(a^2_{i}-a^2_{i-1})=-\sum_{i=1}^{n}\frac{2}{i}(a_{i}a_{i-2}+a_{i-1}a_{i-2})$$
it's
$$a^2_{n}-a^2_{0}=-\sum_{i=1}^{n}\dfrac{2a_{i-2}(a_{i}+a_{i-1})}{i}$$
 A: We can write the equation as $x_n=A_n x_{n-1}$ where $x_n=\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}$ and $A_n=\begin{bmatrix}1&-\delta_n\\ 1&0\end{bmatrix}$ with $\delta_n=2/n$. In the limit $A_n\approx A=\begin{bmatrix}1&0\\ 1&0\end{bmatrix}$, so it makes sense to switch to the basis of eigenvectors of $A$, which are $\begin{bmatrix}0\\1\end{bmatrix}$ and $\begin{bmatrix}1\\1\end{bmatrix}$. Then the transition matrix becomes
$$
A'_n=\begin{bmatrix}\delta_n&\delta_n\\ -\delta_n&1-\delta_n\end{bmatrix}
$$
and the Hilbert-Schmidt norm of $A'_n$ is $\sqrt{1-2\delta_n+4\delta_n^2}=1-\delta_n+O(\delta_n^2)$. Then, since the norm of the product does not exceed the product of the norms, we get $\|\prod_{k=1}^n A'_k\|=O(n^{-2})$. The rest should be clear.
A: Let $$b_n=(n^2-3n)a_n-(2n-4)a_{n-1}$$
Then $b_n=b_{n-1}$
$$a_n=\frac{2n-4}{n^2-3n}a_{n-1}+\frac1{n^2-3n}b$$
Eventually, $|a_n|\lt|a_{n-1}|/2+b/2\lt max(a_{n-1},b)$ so it is bounded above, say by $c$.  The same equation now shows $a_n=O(2c/n)$.  Yet again, the same equation shows $a_n=O(4c/n^2+b/n^2)$, so it is absolutely summable.
A: By considering the OGF
$$ f(x)=\sum_{n\geq 0}a_n x^n $$
the recurrence relation $$ a_{n+2}=a_{n+1}-\frac{2a_n}{n+2} $$
can be written as 
$$ \frac{f(x)-a_0-a_1 x}{x^2} = \frac{f(x)-a_0}{x}-\frac{2}{x^2}\int_{0}^{x}z\,f(z)\,dz $$
or as
$$ f(x)-a_0-a_1 x = x\,f(x)-a_0 x-2\int_{0}^{x} z f(z)\,dz, $$
$$ f'(x)-a_1 = f(x)+x\,f'(x)-a_0 - 2x\,f(x), $$
$$ (1-x) f'(x)- (1-2x) f(x) + (a_0 - a_1) = 0.$$
It follows that $f(x)$ is a linear combination of an entire function of the $(a+bx)e^{c+dx}$ kind and 
$$ g(x)=(1-x) e^{2x} \text{Ei}(2(1-x)) $$
where $\text{Ei}$ is the exponential integral. The coefficients of the Maclaurin series of $g(x)$ are positive from some point on, they decay faster than $\frac{K}{n^2}$ and $\lim_{x\to 1^-}g(x)=0$, hence the coefficients of the Maclaurin series of $f(x)$ are absolutely summable.
