How would one go about computing the following?:
$$\sum_{n=0}^\infty (.5)^{2n+1} \cdot \frac{{2n}\choose{n}}{n+1}$$
The motivation is that this gives the probability that a random walk on a number line will hit the point $0$ given that we move with probability $.5$ left or right starting at $1$. Wolfram tells me it's $1$, how does one prove it?
In fact, let's go further while we're at it! How can one compute the following?:
$$\sum_{n=0}^\infty (.5)^{2n+1} \cdot \frac{{2n}\choose{n}}{n+1} \cdot (2n+1)$$
This gives us the expected time the aformentioned random walk would take (at the rate of one second per move).
EDIT: Hmm... by the comparison test, it seems like the second summation diverges... therefore, the walk would take an infinite amount of time?
The binomial theorem gives the expansion $$ (1-x)^{-1/2}=\sum_{n=0}^\infty\binom{2n}{n}\frac{x^n}{4^n}\tag{1} $$ integrating $(1)$ and dividing by $2$ gives $$ 1-(1-x)^{1/2}=\sum_{n=0}^\infty\binom{2n}{n}\frac{x^{n+1}}{(n+1)2^{2n+1}}\tag{2} $$ Plugging $x=1$ into $(2)$ yields $$ \sum_{n=0}^\infty\binom{2n}{n}\frac1{(n+1)2^{2n+1}}=1\tag{3} $$ which verifies the probability distribution.
$$ \begin{align} \sum_{n=0}^\infty\binom{2n}{n}\frac{2n+1}{(n+1)2^{2n+1}} &=\sum_{n=0}^\infty\binom{2n}{n}\frac{2n+2-1}{(n+1)2^{2n+1}}\\ &=\sum_{n=0}^\infty\binom{2n}{n}\frac1{4^n}-\sum_{n=0}^\infty\binom{2n}{n}\frac1{(n+1)2^{2n+1}}\\ &=(1-1)^{-1/2}-\left(1-(1-1)^{1/2}\right)\\[9pt] &=\infty\tag{4} \end{align} $$ Thus, the second series diverges.
More About Divergence
Asymptotically, $$ \binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}\tag{5} $$ Therefore, asymptotically $$ \binom{2n}{n}\frac{2n+1}{(n+1)2^{2n+1}}\sim\frac1{\sqrt{\pi n}}\tag{6} $$ Thus, the second sum diverges.