For proving the convergence of the following series $$\frac{1}{2}^p+\frac{1\times3}{2\times4}^p+\frac{1\times3\times5}{2\times4\times6}^p+.... \text{ for } p>2$$
I try to use Rabe test to establish the convergence (by showing the limit below is greater than 1) $$\lim_{n\to\infty}n(1-(\frac{2n+1}{2n+2})^p)$$
but when I trying to evaluate the limit, I face a dead end. Therefore, I wonder whether anyone can help with this problem or giving me some hints.
By using Wolfram Alpha, the limit equal to $p/2$, which is consistence with convergence of series for $p>2$. Hope this could be useful.
One straightforward way to compute the limit in question would be to use the Taylor approximation for $(1+x)^p$ when $|x| <1$. Indeed, using the binomial series for $(1+x)^p$, where $|x|<1$ we have $$ (1 + x)^p = 1 + px + O(x^2) \tag{1}. $$ Using $(1)$ for $n$ large we get $$ \left( \frac{2n+1}{2n+2} \right)^p = \left( 1 - \frac{1}{2n+2} \right)^p = 1 - p\frac{1}{2n+2} + O\left(\frac{1}{n^2}\right), $$ hence $$ n \left[ 1 - \left( \frac{2n+1}{2n+2} \right)^p \right]= n p \frac{1}{2n+2} + O\left(\frac 1n \right) \to \frac p2, \ \ \text{ as } n \to \infty. $$