Root Test — Question 9

*Question 9 Determine whether $\displaystyle\sum_{n=1}^{\infty}n^{-n}$ converges or diverges. Use the Root Test and explain why the exponent $n$ makes this test especially efficient.

*Question 9 – Solution

We have $a_n=n^{-n}=(1/n)^n$.

$$\sqrt[n]{|a_n|}=\sqrt[n]{\left(\frac1n\right)^n}=\frac1n.$$ Therefore $$L=\lim_{n\to\infty}\frac1n=0<1.$$

The series converges absolutely. Rather than complicating the calculation, the exponent $n$ is exactly canceled by the $n$th root. Moreover, for any $0<r<1$, eventually $1/n<r$, so $n^{-n}<r^n$ and the tail is dominated by a geometric series.