Convergence of infinite series - (k!)/(k^k)?

Question

HI there, just stuck on a question I can't figure out, would like some help.

The original question:
Determine whether the series converges (infinite series, from 1 to infinity)
(k+1)! / (k+1)^k

I know this can be rewritten as an infinite series from 0 to infinity of
k! / k^k

After that, I have no idea how to formally prove that this series converges. I know it of course it does, I just don't know how to prove it.
Help? =) Thanks!

Answer 1

Well, it doesn't translate directly as the sum from 0 to ∞ of k!/k^k. (The power would have to be k + 1 in the denominator.) It is the same as



∞

Σ k (k!/k^k)

k=0



There is an extra factor of k. If you use the ratio test, calling the kth term a_k



a_(k+1)/a_k = [(k+1)/k] k^k/(k+1)^k = [(k+1)/k] (1 + 1/k)^(-k).



If you take the limit as k->∞, the factor (k+1)/k goes to 1. The remaining factor has limit



lim (1 + 1/k)^(-k) = 1/e.

k->∞



So taken together, you have



lim a_(k+1)/a_k = 1/e.

k->∞



Since this number is less than 1, the series converges by the ratio test.

Answer 2

Σ (k+1)! / (k+1)^k



Apply the ratio test:



lim k→∞ | ((k+2)! / (k+2)^(k+1)) / ((k+1)! / (k+1)^k) |

lim k→∞ | [(k+2)! (k+1)^k] / [(k+1)! (k+2)^(k+1)] |

lim k→∞ | [(k+2)(k+1)k! (k+1)^k] / [(k+1)k!(k+2)(k+2)^k] |



(k+2), (k+1), and k! in the numerator/denominator cancel out.



lim k→∞ | (k+1)^k / (k+2)^k | = L



lim k→∞ ln( |(k+1)^k / (k+2)^k| ) = ln(L)

using log properties

lim k→∞ (k ln( (k+1)/(k+2) ) = ln(L)

lim k→∞ (k ln( (1+1/k)/(1+2/k) ) = ln(L)

lim k→∞ ( ln((1+1/k)/(1+2/k)) / (1/k) ) = 0/0



Apply L'Hopital's rule:

lim k→∞ [ ((k+2) - (k+1)) / (k+2)²) / (-1/k²) ]

lim k→∞ [ (1/(k+2)²) / (-1/k²) ]

lim k→∞ [ -k² / (k² + 4k + 4) ]

lim k→∞ [ -1 / (1 + 4/k + 4/k²) ] = -1 = ln(L)



ln(L) = -1

e^(-1) = L

1/e = L



With the ratio test, if L<1, the series converges absolutely; if L>1, the series diverges.



1/e < 1, so the series converges by the ratio test.

Answer 3

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