Find the values of p that makes the following series converge: $$ \Bigl(\frac{1}{2}\Bigl)^p + \Bigl(\frac{1{}\cdot{}3}{2{}\cdot{}4}\Bigl)^p + \Bigl(\frac{1{}\cdot{}3{}\cdot{}5}{2{}\cdot{}4{}\cdot{}6}\Bigl)^p + \ldots $$
I tried to use Raabe's test but couldn't find $\lim_{n\to\infty} {n\biggl(\Bigl(\frac{2n+2}{2n+1}\Bigl)^p - 1}\biggl)$
If you need to find $\lim_{n\to\infty} {n\biggl(\Bigl(\frac{2n+2}{2n+1}\Bigl)^p - 1}\biggl)$ you can put
$\dfrac{2n+2}{2n+1}=1+\dfrac{1}{2n+1}=1+t\iff n=\dfrac{1-t}{2t}$ then your expression becomes $$\lim_{t\to0}\left(\dfrac{1-t}{2t}\right)((1+t)^p-1)=\lim_{t\to0} \dfrac{(1+t)^p-1}{\frac{2t}{1-t}}$$ You can apply now Hôpital's Rule so you can find
$$\lim_{n\to\infty} {n\biggl(\Bigl(\frac{2n+2}{2n+1}\Bigl)^p - 1}\biggl)=\dfrac p2$$