Divergence or convergence of a recurrence sequence using differential equations We have a recursive sequence $y_{n+1} = \sqrt{\frac{n+3}{n+1}} y_{n}$
for which value of $y_0$ does it converge?
I found this question in a set of Differential Equation problems. I don't have any idea how this can be solved using differential equations. The answer is probably only $0$ but I don't know how to solve this using Differential Equations.
 A: $$\begin{align}
y_{n+1}
&=y_{n}\sqrt{\frac{n+3}{n+1}}\\
&=y_n\sqrt{1+\frac{2}{n+1}}\\
\end{align}$$
So we can find $\lim_{n\to\infty}y_n$ by writing
$$\lim_{n\to\infty}y_n = y_0\prod_{n=1}^\infty \sqrt{1+\frac2{n+1}}=y_0\sqrt{\prod_{n=1}^\infty \left(1+\frac2{n+1}\right)}$$
But the product
$$\prod_{n=1}^\infty \left(1+\frac2{n+1}\right)=\prod_{n=1}^\infty \left(\frac{n+3}{n+1}\right)=\left(\frac42\right)\left(\frac53\right)\left(\frac64\right)\left(\frac75\right)\dots=\frac{(n+2)(n+3)}6\to\infty$$
So one can conclude that
$$\lim_{n\to\infty}y_n=
\begin{cases}
\infty &y_0\gt0\\
-\infty &y_0\lt0\\
0&y_0=0
\end{cases}$$
A: This recurrence relation can be solved with standard techniques. The result is 
$$y_n = \sqrt{\frac{(n+1)(n+2)}{2}}y_0,$$
which clearly diverges for any $y_0\ne 0$. 
The connection to differential equations is the following: 
Rewrite the recurrence as 
$$y_{n+1}-y_n = \left(\sqrt{\frac{n+3}{n+1}}-1\right)y_n.$$
This corresponds to the differential equation 
$$y'(x) = \left(\sqrt{\frac{x+3}{x+1}}-1\right)y(x).$$
This can also be solved explicitly. 
The result is somewhat complicated but for large $x$ we find 
$y(x)\simeq c y_0 x,$ 
where $c$ is a (calculable) nonzero real number. 
This can be easily motivated. 
For large $x$ we have 
$$y'(x) \simeq \frac{1}{x}y(x),$$
and so $y(x)\simeq \mathrm{(const)}y_0 x$. 
More about the connection between recurrence relations and differential equations can be found here. 
