3
$\begingroup$

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:

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

Thanks in advance!

$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Oups... sorry for my incorrect comment that I’ll delete! $\endgroup$ – mathcounterexamples.net Aug 15 '19 at 10:19
  • $\begingroup$ A nice use of Raabe's test (+1). $\endgroup$ – Olivier Oloa Aug 15 '19 at 17:55
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.