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I am studying for a calculus final and have come across this practice question:

$\text{Determine whether the series is absolutely convergent:}$

$\sum\limits_{n=1}^{\infty} \frac{(-1)^n [1 \cdot 3 \cdot 5 \cdot \dotsm \cdot (2n - 1)]}{(2n - 1)!}$

So far, I know I need to use the Ratio Test and have worked my way to the following step:

$\lim_{n \rightarrow \infty} \left\lvert\frac{(2n + 1)(2n - 1)!)}{(2n + 1)!(2n - 1)} \right\rvert$

I know by looking at the answer key that the answer is that indeed the series converges. This implies that the result is less than one.

How can I simplify my step to achieve such an answer?

Any help or hints are greatly appreciated! Thanks!

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  • $\begingroup$ The convergence of a series does not imply that the limit of the ratios of successive terms tends to a number less than $1$. $\endgroup$ – David Mitra Apr 8 '14 at 19:24
  • $\begingroup$ I stated that the convergence of a series imply that the limit of the ratios of successive terms tends to a value less than 1. $\endgroup$ – wonggr Apr 8 '14 at 19:25
  • $\begingroup$ Fixed "typo".${}$ $\endgroup$ – David Mitra Apr 8 '14 at 19:26
  • $\begingroup$ Hint: The numerator of the summand contains all odd numbers from $1$ to $2n-1$ while the denominator contains all number from $1$ to $2n-1$. $\endgroup$ – user10444 Apr 8 '14 at 19:28
  • $\begingroup$ The $2n-1$ at the bottom is I think not correct. (The $(2n-1)!$ is fine.) $\endgroup$ – André Nicolas Apr 8 '14 at 19:33
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I think this is what you're asking:

$ \lim_{n\to\infty}|\frac {(2n+1)(2n-1)!}{(2n+1)!(2n-1)}|= \lim_{n\to\infty}|\frac {(2n-2)!}{(2n)!}|= \lim_{n\to\infty}|\frac {1}{(2n)(2n-1)}|=0$

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  • $\begingroup$ How did you simplify the first step to the second? $\endgroup$ – wonggr Apr 8 '14 at 19:45
  • $\begingroup$ Basically $\frac {2n+1}{(2n+1)!}=\frac {1}{(2n)!}$ and $\frac {(2n-1)!}{(2n-1)}=(2n-2)!$ $\endgroup$ – Ellya Apr 8 '14 at 19:49
  • $\begingroup$ Alright, I got that. Now how about the second step to third? $\endgroup$ – wonggr Apr 8 '14 at 20:04
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Hint:

Recall that $(n+2)! = (n+2)\cdot(n+1)\cdot n!$

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  • $\begingroup$ What about $(n - 2)!$ $\endgroup$ – wonggr Apr 8 '14 at 19:33
  • $\begingroup$ it may come as a surprise but is $(n-2)! = (n-2)(n-3)(n-4)!$ $\endgroup$ – Ant Apr 8 '14 at 19:37
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Try first some algebraic simplification:

$$\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{(2n-1)!}=\frac{1\cdot2\cdot3\cdot4\cdot\ldots\cdot(2n-1)(2n)}{2\cdot4\cdot6\cdot\ldots\cdot(2n)(2n-1)!}=\frac{(2n)!}{2^nn!(2n-1)!}=\frac1{2^{n-1}(n-1)!}$$

and now it looks, imo, much simpler:

$$\frac{a_{n+1}}{a_n}=\frac{2^{n-1}(n-1)!}{2^nn!}=\frac1{2n}\xrightarrow[n\to\infty]{}0\ldots$$

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