## Calculus 8th Edition

$\Sigma_{n=1}^{\infty}\frac{5^{k}}{3^{k}+4^{k}}\gt \Sigma_{n=1}^{\infty}\frac{5^{k}}{4^{k}+4^{k}}$ $=\Sigma_{n=1}^{\infty}\frac{5}{8} (\frac{5}{4})^{k-1}$ A geometric series with common ration $r=\frac{5}{2}$, hence diverging.