# How do you test the series Sigma (2^n(n!))/n^n from n=[1,oo) by the ratio test?

Jan 21, 2018

The series converges.

See below.

#### Explanation:

We have:

sum_(n=1)^oo(2^n(n!))/(n^n)

The ratio test tells us the series will converge if :

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid < 1$

In this case: a_n = (2^n(n!))/(n^n)

So:

lim_(n->oo)abs(a_(n+1)/a_n)=lim_(n->oo)abs((2^(n+1)(n+1)!)/((n+1)^(n+1)))/abs((2^(n)(n)!)/(n^n))

=lim_(n->oo)abs((n^n2^(n+1)(n+1)!)/((n+1)^(n+1)2^n n!))

We can do a bit of cancelling here:

Note that: ${2}^{n + 1} / {2}^{n} = {2}^{n + 1 - n} = 2$

And ((n+1)!)/(n!)=((n+1)timesntimes...times3times2times1)/(ntimes...times3times2times1)

$= \frac{\left(n + 1\right) \times \cancel{n \times \ldots \times 3 \times 2 \times 1}}{\cancel{n \times \ldots \times 3 \times 2 \times 1}} = n + 1$

So we can simplify the limit a bit:

=lim_(n->oo)abs((n^ncolor(blue)(2^(n+1))color(green)((n+1)!))/((n+1)^(n+1)color(blue)(2^n) color(green)(n!)))

$= {\lim}_{n \to \infty} \left\mid \frac{{n}^{n} \textcolor{b l u e}{2} \textcolor{g r e e n}{\left(n + 1\right)}}{{\left(n + 1\right)}^{n + 1}} \right\mid = {\lim}_{n \to \infty} \left\mid 2 \frac{{n}^{n} \left(n + 1\right)}{{\left(n + 1\right)}^{n + 1}} \right\mid$

We can now cancel the $n + 1$ on the top with the power of $n + 1$ on the bottom like so:

${\lim}_{n \to \infty} \left\mid 2 \frac{{n}^{n} \textcolor{red}{\left(n + 1\right)}}{{\left(n + 1\right)}^{n + \textcolor{red}{1}}} \right\mid = {\lim}_{n \to \infty} \left\mid 2 \frac{{n}^{n}}{{\left(n + 1\right)}^{n}} \right\mid$

Obviously for integers $n$ and $n > 0$ this will always be real and positive so there is no need for the "absolute" brackets;

$= 2 {\lim}_{n \to \infty} \frac{{n}^{n}}{{\left(n + 1\right)}^{n}}$

To evaluate this limit consider:

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

So:

$\ln \left(L\right) = \ln \left({\lim}_{n \to \infty} \frac{{n}^{n}}{{\left(n + 1\right)}^{n}}\right) = {\lim}_{n \to \infty} \ln \left({\left(\frac{n}{n + 1}\right)}^{n}\right)$

${\lim}_{n \to \infty} n \ln \left(\frac{n}{n + 1}\right) = {\lim}_{n \to \infty} - n \ln \left(\frac{n + 1}{n}\right)$

$= {\lim}_{n \to \infty} - n \ln \left(1 + \frac{1}{n}\right) = - {\lim}_{n \to \infty} \ln \frac{1 + \frac{1}{n}}{\left(\frac{1}{n}\right)} \to \frac{0}{0}$

So use L'Hopital's rule:

$\frac{d}{\mathrm{dn}} \ln \left(1 + \frac{1}{n}\right) = \frac{1}{1 + \frac{1}{n}} \cdot \left(- \frac{1}{n} ^ 2\right)$

$\frac{d}{\mathrm{dn}} \left(\frac{1}{n}\right) = - \frac{1}{n} ^ 2$

So limit now becomes:

$- {\lim}_{n \to \infty} \ln \frac{1 + \frac{1}{n}}{\left(\frac{1}{n}\right)} = - {\lim}_{n \to \infty} \frac{\frac{1}{1 + \frac{1}{n}} \cdot \left(- \frac{1}{n} ^ 2\right)}{- \frac{1}{n} ^ 2}$

The factor of $\left(- \frac{1}{n} ^ 2\right)$ cancels to give:

$- {\lim}_{n \to \infty} \frac{1}{1 + \frac{1}{n}} = - \frac{1}{1 + 0} = - 1$

So $\ln \left(L\right) = - 1$ so it follows that:

$L = {\lim}_{n \to \infty} \frac{{n}^{n}}{{\left(n + 1\right)}^{n}} = {e}^{- 1} = \frac{1}{e}$

Hence:

$2 {\lim}_{n \to \infty} \frac{{n}^{n}}{{\left(n + 1\right)}^{n}} = \frac{2}{e}$

Finally, it follows that:

lim_(n->oo)abs(a_(n+1)/a_n)=lim_(n->oo)abs((n^n2^(n+1)(n+1)!)/((n+1)^(n+1)2^n n!))=2/e<1

So by the ratio test the series converges.