Computing the sum of a Catalan sequence-- Random-walk motivated 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?
 A: 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.
A: The generating function for Catalan numbers is:
$$\frac{2}{1+\sqrt{1-4x}}=\sum_{i=0}^\infty C_ix^n$$
Can you do the rest? Hint, plug in $x^2$ first. To derive the generating function, show that 
$$C_{n+1}=\sum_{i=1}^nC_kC_{n-k}$$
The second sum diverges.
