Show that the series below converges: $$\sum\limits_{k=1}^{\infty }\frac{1}{k^{k}}=1+\frac{1}{2^{2}}+\frac{1}{3^{3}} +\cdots +\frac{1}{n^{n}}+\cdots$$

Solution We know that the geometric series $\sum\limits_{k=1}^{\infty }\dfrac{1}{2^{k}}$ converges since $\vert r\vert =\dfrac{1}{2}<1$. Now, since $\dfrac{1}{n^{n}}\leq \dfrac{1}{2^{n}}$ for all $n\geq 2$, except for the first term, each term of the series $\sum\limits_{k=1}^{\infty }\dfrac{1}{k^{k}}$ is less than or equal to the corresponding term of the convergent geometric series. So, by the Comparison Test for Convergence, the series $\sum\limits_{k=1}^{\infty }\dfrac{1}{k^{k}}$ converges.