Problem
Determine the convergence or divergence of $\sum\limits_{n =1}^{\infty} \left(n^{\frac{1}{n^2+2n+1}}-1\right). $
Solution
Notice that $$n^{\frac{1}{n^2+2n+1}}-1\leq n^{\frac{1}{n^2}}-1=\exp\left(\frac{\ln n}{n^2}\right)-1\sim \frac{\ln n}{n^2}.$$ Since $\sum\limits_{n =1}^{\infty}n^{-{\frac{3}{2}}}$ is convergent, and $$\dfrac{\dfrac{\ln n}{n^2}}{n^{-\frac{3}{2}}}=\frac{\ln n}{\sqrt{n}}\to 0(n \to \infty),$$ by the comparison test, $\sum\limits_{n =1}^{\infty}\dfrac{\ln n}{n^2}$ converges. Further, so does $\sum\limits_{n =1}^{\infty} \left(n^{\frac{1}{n^2+2n+1}}-1\right)$.