Ratio Test — Question 9

Question 9 – Solution

Step 1: Form the ratio.

$$\begin{align*} \frac{a_{n+1}}{a_n}&=\frac{(n+1)^{n+1}}{((n+1)!)^2}\frac{(n!)^2}{n^n}=\frac{(n+1)^{n-1}}{n^n}\\&=\frac1n\left(1+\frac1n\right)^{n-1}. \end{align*}$$

Step 2: Separate the limiting factors.

We have $(1+1/n)^{n-1}\to e$ and $1/n\to0$. Therefore $L=0$.

Conclusion.

Since $L=0<1$, the positive-term series converges. The simplified ratio resolves the competition between $n^n$ and $(n!)^2$ without guesswork.