Question

Test the convergence of the following series?

Answer 1

The series:

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

is convergent for `x < 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)`

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

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

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

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

Then:

`lim_(n-> oo) abs(a_(n+1)/a_n ) = lim_(n-> oo)( 1+1/n)^n abs x =absx e`

It follows that the series is convergent for `absx < 1/e` and divergent for `abs x > 1/e`

For `abs x = 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->oo) a_n/(1/sqrtn ) = 1/sqrt(2pi)`

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

`sum 1/sqrtn`

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

In conclusion the series:

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

is convergent for `x < 1/e`

Answer 2

The series converges for `color(red)(x < 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->oo)abs((a_(n+1))/(a_n))`

`=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->oo) ((n+1)^n)/(n^n)absx`

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

This is an indeterminate limit with the form `1^oo`. 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->oo)((n+1)/n)^n`

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

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

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

The limit in the exponent is now in the form `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->oo)abs((a_(n+1))/(a_n))=eabsx`

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

`eabsx < 1`

`absx < 1/e`

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

`color(red)(x < 1/e)`