Does someone know how to evaluate the convergence of this series? I started by using the ratio test and setting $a_n = (4^{n-1}+2^n)/(5^{n+1})$, but after that step when rewriting I'm not really able to get rid of the $n$s.
Would really appreciate some help, thanks!
/Nick
You may also try comparison test:$$\frac{4^{n-1}+2^n}{5^{n+1}} \lt \dfrac{4^{n+1}+4^{n+1}}{5^{n+1}}=2\left(\dfrac{4}{5}\right)^{n+1}$$
This is a simple linear combination of geometric series. We have $$\sum_{n=0}^{\infty}\frac{4^{n-1}+2^n}{5^{n+1}}=\frac{1}{20}\sum_{n=0}^{\infty}(\frac{4}{5})^n+\frac{1}{5}\sum_{n=0}^{\infty}(\frac{2}{5})^n=\frac{1}{20}\frac{1}{1-\frac{4}{5}}+\frac{1}{5}\frac{1}{1-\frac{2}{5}}=\frac{7}{12}$$
If you want to use the ratio test, set $$a_n=\frac{4^{n-1}+2^n}{5^{n+1}}.$$ Then $$\frac{a_{n+1}}{a_n}=\frac15\left(\frac{4^n+2^{n+1}}{4^{n-1}+2^n}\right) =\frac45\left(\frac{1+2^{1-n}}{1+2^{2-n}}\right)\to\frac45$$ as $n\to\infty$.
So the ratio test works fine, but the other contributors' solutions are slicker.