Question

Could someone please explain how the sum of k!/k^k from 1 to infinity converges?

Answer 1

oh using the ratio test please

Answer 2

You have to use L'Hopitals

Answer 3

To calculate the limit as k->oo of the ratio.

Answer 4

\[\large \frac{(k+1)!/(k+1)^{(k+1)}}{k!/k^k}=\frac{(k+1)!}{k!}\frac{k^k}{(k+1)^{(k+1)}}\]\[\large =(k+1)\frac{k^k}{(k+1)(k+1)^k}=\frac{k^k}{(k+1)^k}=\left ( \frac{k}{k+1}\right ) ^k\]Then use L'Hopitals to calculate the limit.

Answer 5

Actually, an even easier way using the definition of e^x...

Answer 6

\[\frac{k}{k+1}=1-\frac{1}{k+1}\]\[e^x=\lim_{k \rightarrow \infty} \left ( 1+\frac{x}{k}\right ) ^k\]Then do the math.