show this sum is $\sum_{n=0}^{+\infty}|a_{n}|<+\infty$

Question

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 $$a^2_{n}-a^2_{0}=-\sum_{i=1}^{n}\dfrac{2a_{i-2}(a_{i}+a_{i-1})}{i}$$

Answer 1

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.

Answer 2

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.

Answer 3

Using Empy2's idea, we have $b_n=b_{n-1}$ where $$b_n=(n^2-3n)a_n-(2n-4)a_{n-1}$$ from which $b_n=b_1,$ i.e. $$(n^2-3n)a_n-(2n-4)a_{n-1}=-2a_1+2a_0$$ follows.

Multiplying the both sides by $\dfrac{(n-1)(n-4)!}{2^{n}}$ and setting $c_n=\dfrac{n!}{2^n(n-2)}a_n$, we get $$\begin{align}c_n-c_{n-1}&=\frac{(n-1)(n-4)!}{2^{n-1}}(a_0-a_1) \\\\&=\left(\frac{(n-3)!}{2^{n-1}}-\frac{(n-4)!}{2^{n-2}}\right)(a_0-a_1)+\frac{(n-4)!}{2^{n-3}}(a_0-a_1)\end{align}$$

For $n\ge 4$, we get, setting $f(n)=\displaystyle\sum_{k=4}^{n}\frac{(k-4)!}{2^{k-3}}$,

$$\begin{align}c_n&=c_3+(a_0-a_1)\sum_{k=4}^{n}\left(\frac{(k-3)!}{2^{k-1}}-\frac{(k-4)!}{2^{k-2}}\right)+(a_0-a_1)f(n) \\\\&=\frac{1}{4}a_1-\frac 34a_0+(a_0-a_1)\left(\frac{(n-3)!}{2^{n-1}}-\frac 14\right)+(a_0-a_1)f(n) \\\\&=\left(-1+\frac{(n-3)!}{2^{n-1}}+f(n)\right)a_0+\left(\frac 12-\frac{(n-3)!}{2^{n-1}}-f(n)\right)a_1\end{align}$$

So, we get, for $n\ge 4$, $$\begin{align}a_n&=\frac{2^n(n-2)}{n!}c_n \\\\&=\frac{2^n(n-2)}{n!}\left(-1+\frac{(n-3)!}{2^{n-1}}+f(n)\right)a_0 \\&\qquad\quad+\frac{2^n(n-2)}{n!}\left(\frac 12-\frac{(n-3)!}{2^{n-1}}-f(n)\right)a_1\end{align}$$

Therefore, we have

$$\begin{align}\sum_{n=0}^{\infty}|a_n|&\le |a_0|+|a_1|+|a_2|+|a_3|+\sum_{n=4}^{\infty}\bigg|\frac{2^n(n-2)}{n!}\left(-1+\frac{(n-3)!}{2^{n-1}}+f(n)\right)a_0\bigg| \\\\&\qquad\quad +\sum_{n=4}^{\infty}\bigg|\frac{2^n(n-2)}{n!}\left(\frac 12-\frac{(n-3)!}{2^{n-1}}-f(n)\right)a_1\bigg| \\\\&\le |a_0|+|a_1|+|a_2|+|a_3|+\sum_{n=4}^{\infty}\bigg|-\frac{2^n(n-2)}{n!}a_0\bigg| +\sum_{n=4}^{\infty}\bigg|\frac{(n-3)!}{2^{n-1}}\cdot\frac{2^n(n-2)}{n!}a_0\bigg| \\&\qquad\quad +\sum_{n=4}^{\infty}\bigg|\frac{2^n(n-2)}{n!}a_0f(n)\bigg| +\sum_{n=4}^{\infty}\bigg|\frac 12\cdot \frac{2^n(n-2)}{n!}a_1\bigg| \\&\qquad\quad+\sum_{n=4}^{\infty}\bigg|-\frac{(n-3)!}{2^{n-1}}\cdot\frac{2^n(n-2)}{n!}a_1\bigg|+\sum_{n=4}^{\infty}\bigg|-\frac{2^n(n-2)}{n!}a_1f(n)\bigg| \\\\&\le |a_0|+|a_1|+|a_2|+|a_3|+|a_0|\sum_{n=4}^{\infty}\frac{2^n(n-2)}{n!}+2|a_0|\sum_{n=4}^{\infty}\left(\frac{1}{n-1}-\frac 1n\right) \\&\qquad\quad +|a_0|\sum_{n=4}^{\infty}\frac{2^n(n-2)}{n!}f(n)+|a_1|\sum_{n=4}^{\infty}\frac{2^{n-1}(n-2)}{n!} \\&\qquad\quad +2|a_1|\sum_{n=4}^{\infty}\left(\frac{1}{n-1}-\frac 1n\right) +|a_1|\sum_{n=4}^{\infty}\frac{2^n(n-2)}{n!}f(n) \\\\&\le \frac{11}3|a_0|+3|a_1|+(2|a_0|+|a_1|)\sum_{n=1}^{\infty}\frac{2^{n+2}(n+1)}{(n+3)!} \\&\qquad\quad +(|a_0|+|a_1|)\sum_{n=1}^{\infty}\frac{2^{n+3}(n+1)}{(n+3)!}f(n+3)\tag1\end{align}$$

Now, since $$\lim_{n\to\infty}\left(\frac{2^{n+3}(n+2)}{(n+4)!}\div \frac{2^{n+2}(n+1)}{(n+3)!}\right)=\lim_{n\to\infty}\frac{\frac 2n+\frac{4}{n^2}}{1+\frac 5n+\frac{4}{n^2}}=0$$ we see that $\displaystyle\sum_{n=1}^{\infty}\frac{2^{n+2}(n+1)}{(n+3)!} $ converges by d'Alembert's ratio test.

Also, since

$$\begin{align}&n\left(1-\frac{2^{n+4}(n+2)}{(n+4)!}f(n+4)\div \frac{2^{n+3}(n+1)}{(n+3)!}f(n+3)\right) \\\\&=n\left(1-\frac{2(n+2)}{(n+4)(n+1)}\left(\frac{\frac{n!}{2^{n+1}}}{\sum_{k=4}^{n+3}\frac{(k-4)!}{2^{k-3}}}+1\right)\right) \\\\&\ge n\left(1-\frac{2(n+2)}{(n+4)(n+1)}\left(\frac{\frac{n!}{2^{n+1}}}{\frac{(n-1)!}{2^n}+\frac{(n-2)!}{2^{n-1}}}+1\right)\right) \\\\&=\frac{4+\frac 2n}{1+\frac 5n+\frac{4}{n^2}}\to 4\ \ (n\to\infty)\end{align}$$ we see that $\displaystyle\sum_{n=1}^{\infty}\frac{2^{n+3}(n+1)}{(n+3)!}f(n+3)$ converges by Raabe's test.

It follows from $(1)$ that $$\sum_{n=0}^{\infty}|a_0|\lt \infty$$

Answer 4

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.