I need help with the following excersise:
Evaluate $$\sum_{n=0}^{\infty}{\frac{2^n}{(2n+1){2n\choose n}}}\\\text{Hint: Use identity}\int_0^{\pi/2}{\sin^{2k+1}x\;dx}=\frac{2^{2k}k!^2}{(2k+1)!}$$
My attempt:
$$\sum_{n=0}^{\infty}{\frac{2^n}{(2n+1){2n\choose n}}}=\sum_{n=0}^{\infty}{\frac{2^n}{(2n+1){\frac{2n!}{(2n-n)!n!}}}}\\=\sum_{n=0}^{\infty}{\frac{2^nn!^2}{(2n+1)!}}=\sum_{n=0}^{\infty}{\frac{2^n2^nn!^2}{2^n(2n+1)!}}=\sum_{n=0}^{\infty}{\frac{2^{2n}n!^2}{2^n(2n+1)!}}$$
Applying the identity
$$\sum_{n=0}^{\infty}{\frac{2^{2n}n!^2}{2^n(2n+1)!}}=\sum_{n=0}^\infty{\frac{1}{2^{2n}}\int_0^{\pi/2}{\sin^{2n+1}x\;dx}}$$
And here I am stuck, since I am not sure if I can do any change regarding the sum and integral, any help or tips is helpful. Thanks!
Best Answer
Your final expression has a small error. The equality you intended to write is
$$\sum_{n=0}^\infty \frac{2^n}{(2n+1)\binom{2n}{n}}=\sum_{n=0}^\infty \frac1{2^n}\int_0^{\pi/2}\sin^{2n+1}(x)\,dx$$
Now, if we change the order of the summation and integration (valid by uniform convergence), then we find that
$$\sum_{n=0}^\infty \frac{2^n}{(2n+1)\binom{2n}{n}}=\int_0^{\pi/2}\sin(x)\sum_{n=0}^\infty \frac1{2^n}\left(\sin^{2}(x)\right)^n\,dx$$
Next, sum the geometric series and carry out the resulting integral. Can you wrap this up now?