1. ## convergent or divergent

classify the following series as absolutely convergent, conditionally convergent or divergent:

$
\sum^{\infty}_{n=1} \frac{2.5^{n}n!}{n^n}$

im not sure what test to use for this one, i tried the Ratio test (d'Alembert) but i didn't really get anywhere! Please help!

2. Originally Posted by wik_chick88
classify the following series as absolutely convergent, conditionally convergent or divergent:
$\sum^{\infty}_{n=1} \frac{2.5^{n}n!}{n^n}$
Would it surprise you to learn that $\displaystyle
\sqrt[n]{{\frac{{n!}}
{{n^n }}}} \to \frac{1}
{e}~?$

3. Hello, wik_chick88!

The Ratio Test applies . . . but it takes a lot of fancy footwork.

Classify the following series as absolutely convergent, conditionally
convergent or divergent: . $\displaystyle \sum^{\infty}_{n=1} \frac{2.5^{n}n!}{n^n}$

$\displaystyle R \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{2.5^{n+1}(n+1)!}{(n+1)^{n+1}} \cdot\frac{n^n}{2.5^n\,n!} \;=\;\frac{2.5^{n+1}}{2.5^n} \cdot\frac{(n+1)!}{n!}\cdot\frac{n^n}{(n+1)^{n+1}}$

. . $\displaystyle =\;2.5(n+1)\cdot\frac{n^n}{(n+1)^{n-1}} \;=\;2.5\cdot\frac{n^n}{(n+1)^n} \;=\;2.5\cdot\left(\frac{n}{n+1}\right)^n$

Divide numerator and denominator by $\,n\!:$

. . $\displaystyle R \;=\;2.5 \left(\frac{1}{1+\frac{1}{n}}\right)^n \;=\;\frac{2.5}{(1 + \frac{1}{n})^n}$

$\displaystyle \text{Take the limit: }\;\lim_{x\to\infty}R \;=\;\lim_{x\to\infty} \frac{2.5}{\left(1+\frac{1}{n}\right)^n} \;=\;\frac{2.5}{\underbrace{\lim\left(1+\tfrac{1}{ n}\right)^n}_{\text{This is }e}}$

Therefore: . $\displaystyle \lim_{x\to\infty}R \;=\;\frac{2.5}{e} \;<\;1$

. . The seres converges absolutely.

4. And if we were all as clever as Plato, the root tests would work nicely!