Test the convergence of the following series?

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2 Answers
Apr 14, 2018

Answer:

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#

Apr 14, 2018

Answer:

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)#