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The following series converges? $\displaystyle \sum_{n=0}^{\infty}\frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)(-1)^n }{2 \cdot 4 \cdot 6 \cdot 8 \cdot \dots \cdot (2n)} $ Already concerned with the criteria of reason and Kummer

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What have you tried? What happened when you tried it? Save everyone a lot of work by warning of blind alleys. – Gerry Myerson Nov 29 '12 at 5:22
Can you see that your $n$th term is $(-1)^n\frac{1}{4^n}\binom{2n}{n}$? – alex.jordan Nov 29 '12 at 5:27
Are you sure this starts at n = 0 or should that be n = 1? – Amzoti Nov 29 '12 at 5:29
@Amzoti For $n=0$, the interpretation is that both numerator and denominator are empty products, and equal to $1$. – alex.jordan Nov 29 '12 at 5:30
@Alex: Thanks for the clarification - for some reason that threw me off. – Amzoti Nov 29 '12 at 5:31
up vote 4 down vote accepted

Note that $$\begin{align}\frac{1\cdot3\cdot5\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot(2n)} &=\frac{1\cdot2\cdot3\cdots\cdot(2n)}{(2\cdot4\cdot6\cdot\cdots\cdot(2n))^2}\\ &=\frac{(2n)!}{2^{2n}(1\cdot2\cdot3\cdot\cdots\cdot n)^2}\\ &=\frac{(2n)!}{2^{2n}(n!)^2}\\ \end{align}$$

By Stirling's Approximation for factorials, $$\begin{align}\frac{(2n)!}{2^{2n}(n!)^2} &\leq\frac{e(2n)^{2n+1/2}e^{-2n}}{2^{2n}(\sqrt{2\pi}n^{n+1/2}e^{-n})^2}\\ &=\frac{e(2n)^{2n+1/2}e^{-2n}}{2^{2n}(2\pi)n^{2n+1}e^{-2n}}\\ &=\frac{e(2n)^{1/2}}{(2\pi)n}\\ &=\frac{e(2n)^{1/2}}{(2\pi)n}\\ &=\frac{e}{\pi\sqrt{2n}}\longrightarrow0 \end{align}$$

So you have an alternating series whose terms converge to $0$. Also, $$\frac{1\cdot3\cdot5\cdots\cdot(2n-1)(2n+1)}{2\cdot4\cdot6\cdot\cdots\cdot(2n)(2n+2)}<\frac{1\cdot3\cdot5\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot(2n)}$$ So your terms are decreasing in absolute value. The alternating series test guarantees convergence.

And you can even establish what the series converges to. $$\begin{align} \sum_{n=0}^{\infty}(-1)^n\frac{(2n)!}{2^{2n}(n!)^2} &=\sum_{n=0}^{\infty}(-1)^n\frac{1}{4^n}\binom{2n}{n}x^n\\ &=\sum_{n=0}^{\infty}\binom{-1/2}{n}x^n\\ \end{align}$$ where $x=1$. The change to $\binom{-1/2}{n}$ is easy enough to check, and is nice to know every now and then. This last expression is the powers series for $1/\sqrt{1+x}$. With $x=1$, we are right on the boundary of convergence. But we have already established that the series converges with $x=1$. By a theorem of Abel cited at the very beginning of this paper, we may conclude that the series sums to $1/\sqrt{2}$.

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Partial Answer:


$$a_n= \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1) }{2 \cdot 4 \cdot 6 \cdot 8 \cdot \dots \cdot (2n)} \,.$$


$$a_{n}=\frac{2n-1}{2n}a_{n-1} \,,$$ thus $a_n$ is decreasing.

Because of this, there are only two possible outcomes:

Case 1: $\lim_n a_n=0$, in which case you can use the Alternating Series Test.

Case 2: $\lim_n a_n \neq 0$, in which case the series is divergent.

So, basically your question reduces to: Is

$$\lim_n a_n =0 \, ? \,.$$

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In your expressions for the limits, why do you just put $n$ instead of $n\to\infty$? What I mean is $\lim_n a_n$ vs. $\lim_{n\to\infty} a_n$ – Joe Nov 29 '12 at 18:26
@Joe, the $_{\to\infty}$ is taken as understood from the context. – Gerry Myerson Nov 30 '12 at 2:40
@Gerry That's what I presumed. Thanks. – Joe Nov 30 '12 at 2:47

Let $$a_n = \prod_{k=1}^n \frac{2 k - 1}{2 k}.$$ Then $$\arcsin(x) = \sum_{n=0}^{\infty} \frac{a_n}{2n+1}x^{2 n + 1}$$ and in particular the series $\frac{a_n}{2n + 1}$ is summable with sum $\frac{\pi}{2}$. (For an elementary proof see this answer.) Since the series $\frac{1}{2n+1}$ is itself not summable and $a_n$ is a decreasing sequence this implies that $\lim_{n \to \infty} a_n = 0$ and therefore your alternating series is summable.

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