1. **State the problem:** We want to test the convergence of the series $$\sum_{n=1}^\infty \frac{1}{n^n}$$ using the ratio test. 2. **Recall the ratio test formula:** For a series $$\sum a_n$$, the ratio test uses the limit $$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$$. - If $$L < 1$$, the series converges absolutely. - If $$L > 1$$ or $$L = \infty$$, the series diverges. - If $$L = 1$$, the test is inconclusive. 3. **Apply the ratio test:** Here, $$a_n = \frac{1}{n^n}$$. Calculate $$\frac{a_{n+1}}{a_n} = \frac{\frac{1}{(n+1)^{n+1}}}{\frac{1}{n^n}} = \frac{n^n}{(n+1)^{n+1}}$$. 4. **Simplify the expression:** $$\frac{n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n (n+1)} = \left( \frac{n}{n+1} \right)^n \cdot \frac{1}{n+1}$$. 5. **Evaluate the limit:** $$L = \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^n \cdot \frac{1}{n+1}$$. Note that $$\left( \frac{n}{n+1} \right)^n = \left( 1 - \frac{1}{n+1} \right)^n \to e^{-1}$$ as $$n \to \infty$$. Also, $$\frac{1}{n+1} \to 0$$. Therefore, $$L = e^{-1} \times 0 = 0$$. 6. **Conclusion:** Since $$L = 0 < 1$$, the series $$\sum_{n=1}^\infty \frac{1}{n^n}$$ converges absolutely by the ratio test.