# 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.

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.
\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 $$\lim_{n\to\infty}y_n= \begin{cases} \infty &y_0\gt0\\ -\infty &y_0\lt0\\ 0&y_0=0 \end{cases}$$