Convergence Divergence ?

• Jul 19th 2012, 01:47 AM
pratique21
Convergence 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 AM
girdav
Re: 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 AM
pratique21
Re: 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 AM
pratique21
Re: 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 AM
girdav
Re: 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 AM
pratique21
Re: 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.