Question

G se the ratio test to determine the convergence or divergence of the series. ∞ n! nn n = 1

Answer 1

Reading the series as


`\displaystyle\sum_(n\ge1)(n!)/(n^n)`

With the summand
`a_n=(n!)/(n^n)`, we have


`\left|(a_(n+1))/(a_n)\right|=\left|(((n+1)!)/((n+1)^(n+1)))/((n!)/(n^n))\right|=(n+1)(n^n)/((n+1)^(n+1))=(n^n)/((n+1)^n)=\left(\frac n{n+1}\right)^n`

Then


`\displaystyle\lim_(n\to\infty)\left(\frac n{n+1}\right)^n=\lim_(n\to\infty)e^{n\ln\frac n{n+1}}=\exp\left(\lim_(n\to\infty)n\ln\frac n{n+1}\right)`

This is a pretty standard limit that utilizes L'Hopital's rule. We can write


`\displaystyle\lim_(n\to\infty)n\ln\frac n{n+1}=\lim_(n\to\infty)\frac{\ln\frac n{n+1}}{\frac1n}\stackrel{\mathrm{LHR}}=\lim_(n\to\infty)\frac{\frac1{n(n+1)}}{-\frac1{n^2}}=-1`

which means


`\displaystyle\lim_(n\to\infty)\left(\frac n{n+1}\right)^n=e^(-1)<1`

and so the series converges by the ratio test.