Using the Comparison Test for Convergence

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.