How to calculate the sum of $\sum\limits_{n=1}^{\infty}\left(\frac{4^{n-1}}{5^{n-1}}+\frac{4}{5^{n-1}}\right)$

Question

I know this series is converges by limit test, but i can't find a way to calculate its sum.

$$\sum\limits_{n=1}^{\infty}\left(\frac{4^{n-1}}{5^{n-1}}+\frac{4}{5^{n-1}}\right)$$ Thanks for helping,Here is the solution.

Solutions : $$\sum\limits_{n=1}^{\infty}\left(\frac{4^{n-1}}{5^{n-1}}+\frac{4}{5^{n-1}}\right)=\sum\limits_{n=1}^{\infty}\left(\frac45\right)^{n-1}+4\sum\limits_{n=1}^{\infty}\left(\frac15\right)^{n-1}$$ First thing to do is taking out the constant terms, which is $$\left(\frac45\right)^{-1},\left(\frac15\right)^{-1}$$ We can see there are two geometric series: $$\sum\limits_{n=1}^{\infty}\left(\frac45\right)^{n}=\frac{\frac45}{1-\frac45}=4$$ Another one is: $$\sum\limits_{n=1}^{\infty}\left(\frac15\right)^{n}=\frac{\frac45}{1-\frac45}=\frac14$$

So,we have $$\sum\limits_{n=1}^{\infty}\left(\frac45\right)^{n-1}+4\sum\limits_{n=1}^{\infty}\left(\frac15\right)^{n-1}=\frac{5}{4}\cdot4+4\cdot5\cdot\frac14$$

$$\sum\limits_{n=1}^{\infty}\left(\frac45\right)^{n-1}+4\sum\limits_{n=1}^{\infty}\left(\frac15\right)^{n-1}=10$$

Answer 1

We can use that

$$\sum\limits_{n=1}^{\infty}\left(\frac{4^{n-1}}{5^{n-1}}+\frac{4}{5^{n-1}}\right)=\sum\limits_{n=1}^{\infty}\left(\frac45\right)^{n-1}+4\sum\limits_{n=1}^{\infty}\left(\frac15\right)^{n-1}$$

Can you conclude using geometric series?