Question

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

Answer 1

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

Explanation:

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`

so 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`.