(EDITED) Show that $\displaystyle\sum_{n=0}^{\infty} {2n \choose n } \frac{2^n}{(n+1)\cdot 3^{2n+1}} =\frac{1}{2}$

Question

I have a series which I have shown converges with the ratio test. I've observed that the partial sums appear to be approaching $\frac{1}{2}$.

How would I go about proving this? I have considered using the Squeeze theorem, but I can't find any information on using that theorem to show convergence to a specific limit, and I currently have some misgivings that it is rigorous.

Thank you in advance for your help!

EDIT: The series of interest is $$\sum_{n=0}^{\infty} {2n \choose n } \frac{2^n}{(n+1)\cdot 3^{2n+1}}$$

Answer 1

Rewrite $$S=\sum_{n=0}^{\infty} {2n \choose n } \frac{2^n}{(n+1)\cdot 3^{2n+1}}=\frac 13\sum_{n=0}^{\infty} {2n \choose n } \frac{x^n}{(n+1)}$$ where $x=\frac 29$ that is to say $$S=\frac 1{3x}\sum_{n=0}^{\infty} {2n \choose n } \frac{x^{n+1}}{(n+1)}$$ $$T=\sum_{n=0}^{\infty} {2n \choose n } \frac{x^{n+1}}{(n+1)}\implies T'=\sum_{n=0}^{\infty} {2n \choose n }x^n =\frac{1}{\sqrt{1-4 x}}$$ $$T=\int \frac{dx}{\sqrt{1-4 x}}=-\frac{1}{2} \sqrt{1-4 x}+C$$ and $$C=\lim_{x\to 0} \, \frac{1}{2}\frac{1}{\sqrt{1-4 x}}=\frac 12$$ All of the above make $$S=\frac 1{3x}\frac{1-\sqrt{1-4 x} }2=\frac{1-\sqrt{1-4 x} }{6x}$$ Now, make $x=\frac 29$.