Why Doesn't This Series Converge?

I don't think the series diverges.

$$ \frac{1 \cdot 3 \cdot 7 \cdots (2n-3)}{2 \cdot 4 \cdot 6 \cdots 2n}$$

$$ = \frac{{2n \choose n}}{(2n-1)4^n} = \frac{1}{(2n-1)\sqrt{\pi n}}(1 + O(\frac{1}{n}))$$

using the approximation

$$\frac{ {2n \choose n}}{4^n} = \frac{1}{\sqrt{\pi n}}(1 + O(\frac{1}{n}))$$

and so the series must converge! Perhaps you have a mistake in your computation?

For a more elementary proof of convergence, see the end of the answer.

I believe you should be able to compute it using the series expansion for

$$\frac{1}{\sqrt{1-x^2}} = \sum_{k=0}^{\infty} \frac{{2k \choose k} x^{2k}}{4^k}$$

(you will need to subtract some terms, divide by $x^2$ and integrate).

---

An elementary proof of convergence.

We will show that

if $\displaystyle S_n = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots 2n}$ then $\displaystyle S_n \le \frac{1}{\sqrt{n+1}}$

This proves the partial term of your series, which is $\displaystyle \sum_{k=2}^{n} \frac{S_k}{2k-1} < \sum_{k=2}^{\infty} \frac{1}{(2k-1)\sqrt{k+1}} < C$ (for some constant $C$), and thus is bounded above. Since the series is monotonically increasing and bounded above, it is convergent.

We will prove that $\displaystyle S_n \le \frac{1}{\sqrt{n+1}}$ by induction on $n$.

For $n=1$ it is clearly true.

Now $S_{n+1} = S_n \frac{2n+1}{2n+2}$

Consider $\displaystyle 1 - \frac{2n+1}{2n+2} = \frac{1}{2n+2} \ge \frac{1}{\sqrt{n+2}(\sqrt{n+2} + \sqrt{n+1})} = \frac{\sqrt{n+2} - \sqrt{n+1}}{\sqrt{n+2}}$

Thus $\displaystyle \frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ and so

$\displaystyle S_{n} \le \frac{1}{\sqrt{n+1}} \Rightarrow S_{n+1} \le \frac{1}{\sqrt{n+2}}$

You can use Taylor approximations here. Note that the ratio between consecutive terms is ${2n - 3 \over 2n} = \exp(\ln(1 - 3/2n)) = \exp(-{3 \over 2n} + O(1/n^2))$. So the product is comparable to $\exp(-{3 \over 2} \sum_{i = 2}^n {1 \over n} + O(1/n))$, which in turn is comparable to $\exp(-{3 \over 2} \ln(n))$ or $n^{-{3 \over 2}}$. Thus the series converges.

Comment: Seeing others computing the actual value... once you know it converges (absolutely) you can use the original power series to get the actual value. For $|x| < 1$, $(1 - x)^{1 \over 2} = 1 - {x \over 2} - {1 \over 2} \sum_{n=2}^{\infty} A_nx^n$, where $A_n$ is the $n$ term of the sum. Since all terms of the sum are positive and it converges for $x = 1$, if you take limits as $x$ goes to 1 you get the sum $S$ we're looking at. In other words, 0 = 1 - 1/2 - ${S \over 2}$ or $S = 1$.

And actually if you think about it.. if the series did diverge, taking limits of the series as $x$ goes to 1 from below would give infinity since all terms are positive. Thus just taking limits of both sides as $x$ goes to 1 gives both convergence and the correct value.

Further to the other answers posted, not only does the sum converge, but you can conclude that it converges precisely because of the way it was derived. As you state, the terms $C_n=\frac{1\cdot3\cdot5\cdot7\cdots(2n-3)}{2^nn!}$ arise as the Taylor expansion of $f(x)=\sqrt{x}$ about $x=1$, $$ \sqrt{1-x}=1-\frac{x}{2}-\sum_{n=2}^\infty C_nx^n. $$ As $\sqrt{1-x}$ is a well-defined analytic function for $\Vert x\Vert < 1$, this has radius of convergence 1. Also, as the square root is well defined at 0, monotone convergence gives $$ \sum_{n=2}^\infty C_n=\lim_{x\to1}\sum_{n=2}^\infty C_nx^n=\lim_{x\to1}\left(1-\frac{x}{2}-\sqrt{1-x}\right)=\frac12. $$