## Calculus (3rd Edition)

We compare the given series $\sum_{n=1}^{\infty} \frac{1}{n^n}$ with the geometric series $\sum_{n=1}^{\infty} \frac{1}{2^{n}}$ which is a convergent series ($r=\frac{1}{2}\lt 1$). Thus, by the comparison test, the given series $\sum_{n=1}^{\infty} \frac{1}{n^n}$ also converges.