Finding sum of geometric series $\sum\limits_{k=5}^ \infty \left(\frac{e}{\pi}\right)^{k-1} $

Question

Determine the sum of the following geometric series.

I got confused here because I'm used to starting with $k=1$ and now suddenly it's $k=5$. Is there any difference?

$$\sum_{k=5}^ \infty \left(\frac{e}{\pi}\right)^{k-1} $$

Answer 1

you can use the geometric series $$\sum_{k=0}^{\infty }x^k=\frac{1}{1-x} \quad{|x|<1}$$ so $$\sum_{k=5}^ \infty (\frac{e}{\pi})^{k-1}=\sum_{k=0}^ \infty (\frac{e}{\pi})^{k+4}=(\frac{e}{\pi})^4\sum_{k=0}^ \infty (\frac{e}{\pi})^{k}=(\frac{e}{\pi})^4\frac{1}{1-\frac{e}{\pi}}$$

Answer 2

Hint: $$\sum_{k=a}^\infty f(k) = \sum_{k=1}^\infty f(k) - \sum_{k=1}^{a-1}f(k)$$ or more generally if $a < b < c$ $$\sum_{k=b}^c f(k) = \sum_{k=a}^c f(k) - \sum_{k=a}^{b-1}f(k)$$