# Math Help - D'Alembert's ratio question

1. ## D'Alembert's ratio question

hi everyone,
Use D'Alembert's ratio test to test the convergent of the following series:

i)1+ $\frac{2^2}{2!}+\frac{3^3}{3!}+\frac{4^4}{4!}$

im stuck, i found the general term is
$\frac{n^n}{n!}$

2. If $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}$ exists and is less than 1, the serie converges, if is great than 1, the serie diverges.

$\frac{a_{n+1}}{a_n}=\dfrac{\frac{(n+1)^{n+1}}{(n+1 )!}}{\frac{n^n}{n!}}= \frac{(n+1)^{n+1}n!}{(n+1)!n^n}=(\frac{n+1}{n})^n$
Appliying the limit, the result is e, then, the serie diverges.

3. thank you felper for helping to clarify, am just wondering, how you get the result e? need some advise on this.

thank you so much

4. Originally Posted by anderson
thank you felper for helping to clarify, am just wondering, how you get the result e? need some advise on this.

thank you so much
It's a standard limit - see

http://en.wikipedia.org/wiki/E_(mathematical_constant)

5. $\lim_{n\to\infty}\left(\dfrac{n+1}{n} \right)^n=\lim_{n\to\infty}\left(1+\dfrac{1}{n} \right)^n=e$

http://en.wikipedia.org/wiki/E_(mathematical_constant)