Originally Posted by

**anderson** Use D'Alembert's ratio test to determine the convergent of the following series(hint:Find the general term first)

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

the general term: $\displaystyle \frac {n^n}{n!}$

so is it:

$\displaystyle \frac {\frac {(n+1)^{(n+1)}}{{n+1}!}}{\frac {n^n}{n!}}$

is it correct? i don't know how to solve the limit...please help me....

ii) $\displaystyle \frac{1}{2} +\frac {2}{3} + \frac{3}{4}+ \frac{4}{5}$

general term

$\displaystyle \frac {n}{n+1}$

is it

$\displaystyle \frac{\frac {n+1}{n+2}}{\frac {n}{n+1}}$

$\displaystyle \frac{n^2+2n+2}{n^2+2n}$

limit = 1

convergent

iii)$\displaystyle \frac{2}{5}+ \frac {2^2}{6}+ \frac {2^3}{7} +\frac {2^4}{8}$

general term : $\displaystyle \frac {2^n}{n+4}$

$\displaystyle \frac{\frac {2^{n+1}}{n+5}}{\frac {2^n}{n+4}}$

$\displaystyle \frac{2n+8}{n+5}$

limit 2

2>1, divergent

please help guide & correct me, all help appreciated..thank you & regards