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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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    You have an alternating series with terms that are decreasing in absolute value. Can you show that the terms converge to $0$? If you apply Stirling's formula, you can see that your terms have absolute value akin to $\frac{1}{\sqrt{n}}$.2012-11-29

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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:

Let

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

Then

$a_{n}=\frac{2n-1}{2n}a_{n-1} \,,$ 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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    @Gerry That's what I presumed. Thanks.2012-11-30
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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.