How would one compute $ \sum_{n=0}^\infty\frac{z^{n-2}}{5^{n+1}} $ where $0\lt|z|\lt5$?
I have literally no idea where to start, all I know is that the answer will not have summations. Any help would be appreciated!
How would one compute $ \sum_{n=0}^\infty\frac{z^{n-2}}{5^{n+1}} $ where $0\lt|z|\lt5$?
I have literally no idea where to start, all I know is that the answer will not have summations. Any help would be appreciated!
HINT:
$\frac{z^{n-2}}{5^{n+1}}=\frac1{5z^2}\left(\frac{z}5\right)^n$
$\sum_{n=0}^\infty {z^{n-2}\over 5^{n+1}}= {1\over 5z^2}\sum_{n=0}^\infty\left({z\over 5}\right)^n$
Does this make it a bit more palatable?
As said above,
$ \sum_{n=0}^\infty {z^{n-2}\over 5^{n+1}}= {1\over 5z^2}\sum_{n=0}^\infty\left({z\over 5}\right)^n. $
To solve this, it may aid you to make the substitution $u=\frac{z}{5}$. Then, $ {1\over 5z^2}\sum_{n=0}^\infty\left({z\over 5}\right)^n=\frac{1}{5z^2}\sum_{n=0}^{\infty}u^n. $
The sum is geometric in $u$; thus, you apply the geometric series formula.