To find Convergence/Divergence of

$\displaystyle u(n) = \sum \frac{1}{n!}$

and of this one

$\displaystyle u(n) =\sum \frac{1.2.3.4.5.6.....}{1.3.5.7.9......} $

Thank you. :)

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- Jul 19th 2012, 01:47 AMpratique21Convergence Divergence ?
To find Convergence/Divergence of

$\displaystyle u(n) = \sum \frac{1}{n!}$

and of this one

$\displaystyle u(n) =\sum \frac{1.2.3.4.5.6.....}{1.3.5.7.9......} $

Thank you. :) - Jul 19th 2012, 02:04 AMgirdavRe: Convergence Divergence ?
You should write $\displaystyle u(n)=\frac 1{n!}$ for the first, not for the whole sum since it won't depend on $\displaystyle n$. Use the fact that $\displaystyle n!\geq n(n-1)$. For the second, could you write the terms with a formula on which $\displaystyle n$ appears?

- Jul 19th 2012, 02:15 AMpratique21Re: Convergence Divergence ?
I do not understand how the fact that n! >= n(n-1) can be used for to solve this .

The second one can be simplified to get :-

$\displaystyle \frac{2(n!)(n!)}{(2n-1)!} $

Enlighten me.

:) - Jul 19th 2012, 02:39 AMpratique21Re: Convergence Divergence ?
It is e itself actually, so it converges !

e= $\displaystyle \sum \frac{1}{n!} $

But how do you prove it ? - Jul 19th 2012, 02:41 AMgirdavRe: Convergence Divergence ?
We have $\displaystyle \frac 1{n!}\leq \frac 1{n(n-1)}$ and the sum of the terms of the RHS is telescopic.

For the second series, first try ratio test. If you can't conclude, try Raabe-Duhamel test. - Jul 19th 2012, 03:02 AMpratique21Re: Convergence Divergence ?
thanks girdav. I shall try it using the tests you mentioned. However this problem was presented before discussion on ratio test or raabe-duhamel test. So i guess we are expected to solve it using comparison test.