1. ## Convergence Divergence ?

To find Convergence/Divergence of
$u(n) = \sum \frac{1}{n!}$

and of this one

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

Thank you.

2. ## Re: Convergence Divergence ?

You should write $u(n)=\frac 1{n!}$ for the first, not for the whole sum since it won't depend on $n$. Use the fact that $n!\geq n(n-1)$. For the second, could you write the terms with a formula on which $n$ appears?

3. ## 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 :-
$\frac{2(n!)(n!)}{(2n-1)!}$

Enlighten me.

4. ## Re: Convergence Divergence ?

It is e itself actually, so it converges !
e= $\sum \frac{1}{n!}$
But how do you prove it ?

5. ## Re: Convergence Divergence ?

We have $\frac 1{n!}\leq \frac 1{n(n-1)}$ 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.

6. ## 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.