Why Doesn't This Series Converge?

I am teaching a Calc II course and came across the following series when finding the interval of convergence for the Taylor series of $f(x)=\sqrt{x}$ centered at $x=1$:

$$\sum_{n=2}^\infty \frac{1\cdot 3\cdot 5\cdot 7\cdots (2n-3)}{2^nn!}.$$

I know that it doesn't converge, but how do you show it? I would like to be able to use only the technology of a Calc II course, but any answer would be enlightening.

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It is probably worth noting that the terms go to $0$. In fact, this would be true even if the denominator stopped at $2n-2$, but here it is easier to see because the term is $\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-3}{2n-2}\cdot\frac{1}{2n}$ – Jonas Meyer Nov 8 '10 at 4:06

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

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

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This is a nice answer. I was still looking for an "elementary" way to show it diverges! Now the question may be whether this can be adapted to "Calc II level". – Jonas Meyer Nov 8 '10 at 4:49
@Jonas: No idea what Calc II entails... Stirling approximation formula is Calc II? In any case, I think there was an inequality with 2n choose n, 4^n and sqrt(n) with an elementary proof. I saw it a long time back, perhaps I can remember it. If I do, I will post it. – Aryabhata Nov 8 '10 at 4:54
@Moron: Thanks. I'm not sure what it entails either in this case, since it varies so much by university and professor, but (unfortunately?) my impression is that Stirling's formula is usually not covered. I certainly don't mean to imply that this takes anything away from the answer, but if you feel inclined to post more, I'll check back :) – Jonas Meyer Nov 8 '10 at 5:03
@Jonas: I understand :-) I will ping you if I find the elementary proof. – Aryabhata Nov 8 '10 at 5:05
Nice. I'd upvote again if I could. – Jonas Meyer Nov 8 '10 at 8:19

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.

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Ah, the point in your final paragraph is what my answer was getting at. Monotone convergence proves that the sequence converges as well as giving the limit. – George Lowther Nov 8 '10 at 18:44

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 $$\sum_{n=2}^\infty C_n=\lim_{x\to1}\sum_{n=2}^\infty C_nx^n=\lim_{x\to1}\left(1-\frac{x}{2}-\sqrt{1-x}\right)=\frac12.$$

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$\displaystyle \sum_{n=1}^\infty \frac{\binom{2n}{n}}{(2n-1)4^n}.$

Note that

$\displaystyle \frac{\binom{2n}{n}}{2n-1} = \frac{(2n)!}{(2n-1)n!n!} = 2 \frac{(2n-2)!}{n!(n-1)!} = 2C_{n-1},$

where $C_n$ is the $n$th Catalan number.

Now $C_n$ counts the number of paths of length $2(n+1)$ that reach the origin for the first time at their end, always being to the right of the origin.

Since a one-dimensional random walk will get back to the origin w.p. $1$,

$\displaystyle \sum_{n=1}^\infty \frac{C_{n-1}}{4^n} = \frac{1}{2}.$

Therefore the original sum equals

$\displaystyle \sum_{n=1}^\infty \frac{2C_{n-1}}{4^n} = 1.$

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Using Stirling formula $n!\sim \sqrt {2\pi n}(n/e)^n$, I also get that the $n$th term of the series is $O(1/n^{3/2})$. So the series is convergent. In the derivation I used that $$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n=e^{-1}.$$

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In fact the sum converges. You can find a closed form by massaging the formula that I posted on Hadamard products viz. $\rm\ (1-4x)^{-1/2} \;=\; \sum\ \binom{2n}{n}\ x^n\:.\$ See also this post. Alternatively:

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Mma knows about !!, so you don't even need to massage the input :) – Mariano Suárez-Alvarez Nov 8 '10 at 4:53