Test the convergence of the following series?

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2 Answers
Apr 14, 2018

The series:

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

is convergent for x < 1/ex<1e

Explanation:

The general term of the series is:

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

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!)))an+1an=∣ ∣ ∣(n+1)n+1xn+1(n+1)!nnxnn!∣ ∣ ∣

abs(a_(n+1)/a_n ) = abs ((n+1)^(n+1)/n^n (n!)/((n+1)!) x^(n+1)/x^n)an+1an=∣ ∣(n+1)n+1nnn!(n+1)!xn+1xn∣ ∣

abs(a_(n+1)/a_n ) = (n+1)^(n+1)/n^n 1/(n+1) abs xan+1an=(n+1)n+1nn1n+1|x|

abs(a_(n+1)/a_n ) = (n+1)^n/n^nabs xan+1an=(n+1)nnn|x|

abs(a_(n+1)/a_n ) =( (n+1)/n)^n abs xan+1an=(n+1n)n|x|

abs(a_(n+1)/a_n ) =( 1+1/n)^n abs xan+1an=(1+1n)n|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

Apr 14, 2018

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)