## Calculus (3rd Edition)

We have the given series $\sum_{n=1}^{\infty} \frac{4^n}{5^n-2n}$ We apply the limit comparison test with $b_n=(\frac{4}{5})^n$ (a convergent geometric series with $r=4/5\lt 1$): $L=\lim_{n\rightarrow\infty} \frac{a_n}{b_n}=\lim_{n\rightarrow\infty}\frac{4^n}{5^n-2n}\times(\frac{5}{4})^n=\lim_{n\rightarrow\infty}\frac{1}{1-\frac{2n}{5^n}}$ The $\frac{2n}{5^n}$ term is indeterminate, so we apply L'Hopital's rule and derive it: $\lim_{n\rightarrow\infty}\frac{2n}{5^n}=\lim_{n\rightarrow\infty}\frac{2}{5^n\ln 5}=0$ Thus the overall limit is: $L=\lim_{n\rightarrow\infty}\frac{1}{1-\frac{2n}{5^n}}=1$ Since $L=1$, our starting series converges.