Home > Calculus > Determining the Radius and Interval of Convergence for a Power Series

What is the interval of convergence of #sum_1^oo (x^n *n^n)/(n!)#?

Answer 1

Evelyn Agard

#(-1/e, 1/e)# or possibly #[-1/e, 1/e]#

Let #a_n = n^n/(n!)#

Then:

#a_(n+1) = (n+1)^(n+1)/((n+1)!) = (n+1)^n/(n!)=((n+1)/n)^n n^n/(n!)#

#=(1+1/n)^n a_n#

Now #(1+1/n)^n -> e# as #n->oo#

So: #(x^(n+1)*(n+1)^(n+1))/((n+1)!) -: (x^n*n^n) / (n!) = (a_(n+1)/a_n)x -> ex# as #n->oo#

Hence if #abs(x) < 1/e# then #sum_(n=0)^oo (x^n*n^n) / (n!)# is absolutely convergent.

If #abs(x) > 1/e# then #EE N in ZZ# such that:

#abs((x^(n+1)*(n+1)^(n+1))/((n+1)!) -: (x^n*n^n) / (n!)) > 1# for all #n >= N#

Consequently, the sum diverges.

I am not sure about the case #abs(x) = 1/e#, since #(1+1/n)^n < e# for all #n in NN#.

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Answer 2

Christopher Agresta

The interval of convergence for the series ( \sum_{n=1}^{\infty} \frac{x^n \cdot n^n}{n!} ) is ( -e^{-1} < x < e^{-1} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.