# Hint for showing $\sum_{n=1}^\infty{\prod_{j=0}^n{(1-\frac{1}{\sqrt{j+2}})}}$ is convergent

I need a hint for showing that the following series is convergent $$\sum_{n=1}^\infty{\prod_{j=0}^n{(1-\frac{1}{\sqrt{j+2}})}}$$

What I have so far:

$$\sum_{n=1}^\infty{\prod_{j=0}^n{(1-\frac{1}{\sqrt{j+2}})}}<\sum_{n=1}^\infty{\prod_{j=0}^n{(1-\frac{1}{\sqrt{n+2}})}}=\sum_{n=1}^\infty{(1-\frac{1}{\sqrt{n+2}})^n}$$

Alternative way for a more general problem: apply Raabe's test with $$a_n=\prod_{j=0}^n\left(1-\frac{1}{(j+2)^{\alpha}}\right).$$ and $$0<\alpha<1$$ (your case is $$\alpha=1/2$$). Then $$n\left(\frac{a_n}{a_{n+1}}+1\right)=n\left(\frac{1}{1-\frac{1}{(n+3)^\alpha}}-1\right)=\frac{n}{(n+3)^\alpha-1}\to +\infty$$ and the series is convergent.

• Oups... sorry for my incorrect comment that I’ll delete! – mathcounterexamples.net Aug 15 '19 at 10:19
• A nice use of Raabe's test (+1). – Olivier Oloa Aug 15 '19 at 17:55

Only a hint as you requested...

Naming

$$P_n = \prod_{j=0}^n{\left(1-\frac{1}{\sqrt{j+2}}\right)},$$

take the logaritm $$\ln P_n$$ and use power series of $$\ln(1-x)=-x -\frac{x^2}{2} -\dots$$ around $$0$$.

You can then use comparison integral test comparison to evaluate each term of the expansion convergence and take the exponential.