1. ## D'Alembert's ratio test

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

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

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

so is it:
$\frac {\frac {(n+1)^{(n+1)}}{{n+1}!}}{\frac {n^n}{n!}}$

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

general term
$\frac {n}{n+1}$
is it

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

$\frac{n^2+2n+2}{n^2+2n}$
limit = 1
convergent

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

general term : $\frac {2^n}{n+4}$
$\frac{\frac {2^{n+1}}{n+5}}{\frac {2^n}{n+4}}$

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

limit 2

2>1, divergent

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

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

so is it:
$\frac {\frac {(n+1)^{(n+1)}}{{n+1}!}}{\frac {n^n}{n!}}$

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

general term
$\frac {n}{n+1}$
is it

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

$\frac{n^2+2n+2}{n^2+2n}$
limit = 1
convergent

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

general term : $\frac {2^n}{n+4}$
$\frac{\frac {2^{n+1}}{n+5}}{\frac {2^n}{n+4}}$

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

limit 2

2>1, divergent

It would be easier to note that $\frac{n^n}{n!}\to\infty$...but $\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}=\lef t(1+\frac{1}{n}\right)^n\to?$. You should recognize this last limit.

is it e?

is 1 >e, the answer is divergent?

4. Originally Posted by anderson

is it e?

is 1 >e, the answer is divergent?

5. Originally Posted by Drexel28
thank you so much for helping. just wanted to know if i did the other two questions correctly, can anyone help me to confirm..really appreciate all your help & guidance.

for ii) the answer is no conclusion because L=1.

thank you & regards

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