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Hi! I am working on some ratio and root test online homework problems for my calc2 class and I am not sure how to completely solve this problem. I guessed on the second part that it converges, but Im not sure how to solve of the value that it converges to. If someone could possibly help me with this problem it would be greatly appreciated.

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You aren't expected to figure out what the value of the series is (although in time you might figure out it has something to do with $e+e^{-1}$). Did you actually compute $\rho$? What is $$\lim_{n\to\infty} \frac{(2n)!}{(2n+2)!}?$$

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$\rho=\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\rightarrow\infty}|\frac{\frac{1}{(2n+2)!}}{\frac{1}{(2n)!}}|=\lim_{n\rightarrow\infty}|\frac{(2n)!}{(2n+2)!}|=\lim_{n\rightarrow\infty}\frac{1}{(2n+1)(2n+2)}=\lim_{n\rightarrow\infty}\frac{1}{4n^2+5n+2}=0$

I think you may have overlooked the fact that it was a factorial?

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  • $\begingroup$ Thank you so much! I was making silly mistakes, but your solution really cleared things up for me and I understand what I was suppose to do now. $\endgroup$ – user124539 Mar 30 '14 at 21:08
  • $\begingroup$ Tis all good, remember the ratio test tells us what is happening term by term, i.e if terms are getting closer to each other or further away, or just staying the same. $\endgroup$ – Ellya Mar 30 '14 at 21:10
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We have $a_n = \frac{1}{(2n)!}$ and $a_{n+1} = \frac{1}{(2n+2)!}$

\begin{align}\lim_{n \rightarrow \infty} \Big|\frac{a_{n+1}}{a_n}\Big| &= \lim_{n \rightarrow \infty} \Bigg|\frac{(2n)!}{(2n+2)!}\Bigg| \\ &= \lim_{n \rightarrow \infty} \Bigg|\frac{(2n)(2n-1)(2n-2)\cdots}{(2n+2)(2n+1)(2n)(2n-1)(2n-2)\cdots}\Bigg| \\ &= \lim_{n \rightarrow \infty} \Bigg|\frac{1}{(2n+2)(2n+1)}\Bigg| \\ &=0 < 1 \end{align}

Hence, by the Ratio Test, the series $\sum_{n=1}^{\infty} \frac{1}{(2n)!}$ is convergent.

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