# Is the series \sum_(n=0)^\infty n^n/(4^n n!) absolutely convergent, conditionally convergent or divergent?

## (Use the appropriate test.)

Apr 20, 2018

Converges absolutely by the Ratio Test.

#### Explanation:

To determine absolute convergence, we use the Ratio Test, which tells us if we have some series $\sum {a}_{n} ,$ we let

$L = {\lim}_{n \to \infty} | {a}_{n + 1} / {a}_{n} |$ and:

$L < 1$ implies absolute convergence

$L > 1$ implies divergence

$L = 1$ means the test is inconclusive and we need to use another.

Here, a_n=(n^n)/(4^n n!)

Then,

a_(n+1)=(n+1)^(n+1)/(4^(n+1)(n+1)!)

Recalling that division is multiplication by the reciprocal:

L=lim_(n->oo)|(n+1)^(n+1)/(4^(n+1)(n+1)!)*(4^n n!)/(n^n)|

(n+1)! = (n+1)(n!) so the factorials will cancel.

${4}^{n} / {4}^{n + 1} = {4}^{n - \left(n + 1\right)} = {4}^{-} 1 = \frac{1}{4}$

=lim_(n->oo)|((n+1)^(n+1)cancel(n!))/(n^n4(n+1)cancel(n!))|

${\left(n + 1\right)}^{n + 1} / \left(n + 1\right) = {\left(n + 1\right)}^{n + 1 - 1} = {\left(n + 1\right)}^{n}$

$= \frac{1}{4} {\lim}_{n \to \infty} {\left(n + 1\right)}^{n} / \left({n}^{n}\right) = \frac{1}{4} < 1$

We've dropped the absolute value bars -- as we go to infinity, everything is positive, so no need for absolute value.

Furthermore, that limit is $1$ as the numerator and denominator are raised to the same power, so when we go to infinity, they grow at an equivalent rate.

We could prove this with l'Hospital's Rule, but differentiating the bottom is extremely tedious, so the geometric argument is fine.

Thus, the series converges absolutely.