Sum (n!x^n / n^n) n=1, infinity

I know the answer is "e", but I'm not sure how the book got the solution.

also, sum (x^(2n) / n!) n=o, infinity

answer: all real solutions

I'm in differentials and I just need to brush up on some calc 2 series'.

You should have mentioned that you are looking for the radius of convergence for each power series.

1) Use the Ratio Test.

r = lim(n→∞) |[(n+1)! x^(n+1) / (n+1)^(n+1)] / [n! x^n / n^n]|
..= |x| * lim(n→∞) 1/[(n+1)^n/n^n]
..= |x| * lim(n→∞) 1/(1 + 1/n)^n
..= |x| * 1/e, by the limit definition for e
..= |x|/e.

Since r = |x|/e, the series converges for r < 1 <==> |x| < e, and
diverges for r > 1 <==> |x| > e. Hence, the radius of convergence equals e.
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2) Similarly,
r = lim(n→∞) |[x^(2n+2) / (n+1)!] / [x^(2n) / n!]|
..= |x|^2 * lim(n→∞) 1/(n+1)
..= |x|^2 * 0
..= 0.

Since r = 0 < 1 for all x, the series converges for all x.
Hence the radius of convergence equals ∞.

I hope this helps!

My head already hurts