# Test the convergence of the following series?

## Apr 14, 2018

The series:

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

is convergent for $x < \frac{1}{e}$

#### Explanation:

The general term of the series is:

a_n = (n^nx^n)/(n!)

Using the ratio test we can evaluate the ratio.

abs(a_(n+1)/a_n ) = abs (( ( ( n+1)^(n+1)x^(n+1))/((n+1)!)) / ( (n^nx^n)/(n!)))

abs(a_(n+1)/a_n ) = abs ((n+1)^(n+1)/n^n (n!)/((n+1)!) x^(n+1)/x^n)

$\left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\left(n + 1\right)}^{n + 1} / {n}^{n} \frac{1}{n + 1} \left\mid x \right\mid$

$\left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\left(n + 1\right)}^{n} / {n}^{n} \left\mid x \right\mid$

$\left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\left(\frac{n + 1}{n}\right)}^{n} \left\mid x \right\mid$

$\left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\left(1 + \frac{1}{n}\right)}^{n} \left\mid x \right\mid$

Then:

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n} \left\mid x \right\mid = \left\mid x \right\mid e$

It follows that the series is convergent for $\left\mid x \right\mid < \frac{1}{e}$ and divergent for $\left\mid x \right\mid > \frac{1}{e}$

For $\left\mid x \right\mid = \frac{1}{e}$ we have:

a_n = 1/(n!)(n/e)^n

Using Stirling's approximation:

n! ~ sqrt(2pin)(n/e)^n

we can see that

a_n ~ 1/sqrt(2pin)

which means:

${\lim}_{n \to \infty} {a}_{n} / \left(\frac{1}{\sqrt{n}}\right) = \frac{1}{\sqrt{2 \pi}}$

As the limit is finite, based on the limit comparison test they have the same character, and the series

$\sum \frac{1}{\sqrt{n}}$

is divergent based on the $p$-series test.

In conclusion the series:

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

is convergent for $x < \frac{1}{e}$

Apr 14, 2018

The series converges for $\textcolor{red}{x < \frac{1}{e}}$

#### Explanation:

We are given the first four terms of a series:

x+(2^2x^2)/(2!)+(3^3x^3)/(3!)+(4^4x^4)/(4!)+cdots=sum_(n=1)^oo(n^nx^n)/(n!)

We can test the convergence of this series using the ratio test.

Let a_n=(n^nx^n)/(n!) and a_(n+1)=((n+1)^(n+1)x^(n+1))/((n+1)!)

Now let's evaluate the limit of the ratio of these two terms as $n$ goes to infinity.

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

=lim_(n->oo) abs((((n+1)^(n+1)x^(n+1))/((n+1)!))/((n^nx^n)/(n!)))

=lim_(n->oo) abs(((n+1)^(n+1)x^(n+1))/((n+1)!)*(n!)/(n^nx^n))

=lim_(n->oo) abs(((n+1)^(n+1)x)/((n+1)n^n)

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

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

This is an indeterminate limit with the form ${1}^{\infty}$. Let's try to use the properties of exponents and logarithms to manipulate it into a form where we can apply L'Hôpital's rule.

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

=e^(lim_(n->oo)ln[((n+1)/n)^n]

=e^(lim_(n->oo)nln[(n+1)/n]

$= {e}^{{\lim}_{n \to \infty} \frac{\ln \left[\frac{n + 1}{n}\right]}{\frac{1}{n}}}$

The limit in the exponent is now in the form $\frac{0}{0}$

Applying L'Hôpital's rule to the exponent (taking the derivative of both the numerator and denominator), we get:

=e^(lim_(n->oo)(n/(n+1)*(n-(n+1))/n^2)/(-1/n^2)

=e^(lim_(n->oo)(n/(n+1)*-1/n^2)/(-1/n^2)

=e^(lim_(n->oo)n/(n+1)

$= {e}^{1}$

$= e$

Therefore our ratio test yields the result:

${\lim}_{n \to \infty} \left\mid \frac{{a}_{n + 1}}{{a}_{n}} \right\mid = e \left\mid x \right\mid$

According to the ratio test, the series will converge if:

$e \left\mid x \right\mid < 1$

$\left\mid x \right\mid < \frac{1}{e}$

But the question states that $x > 0$, so this can be simplified to

$\textcolor{red}{x < \frac{1}{e}}$