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

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?

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

• Just assume that the sum starts at $1$, determine the solution and then subtract the sum from $1$ to $4$ from the result. Du you understand what I mean? – Moritz Sep 12 '16 at 13:49
• determine the solution, yes got it, subtract sum from 1 to 4, didn't get it, could you be more specific – Teeban Sep 12 '16 at 13:51
• $\sum_{k=1}^\infty a_k = \sum_{k=1}^4 a_k + \sum_{k=5}^\infty a_k$ – Nicholas Stull Sep 12 '16 at 13:53
• @Teeban: Look at the hint from Nicholas Stull. – Moritz Sep 12 '16 at 13:54

you can use the geometric series $$\sum_{k=0}^{\infty }x^k=\frac{1}{1-x} \quad{|x|<1}$$ so $$\sum_{k=5}^ \infty (\frac{e}{\pi})^{k-1}=\sum_{k=0}^ \infty (\frac{e}{\pi})^{k+4}=(\frac{e}{\pi})^4\sum_{k=0}^ \infty (\frac{e}{\pi})^{k}=(\frac{e}{\pi})^4\frac{1}{1-\frac{e}{\pi}}$$
• $$\sum_{k=1}^ \infty ar^{k-1}$$ is the formula for sum, if k=0 how can i find the sum – Teeban Sep 12 '16 at 13:48
Hint: $$\sum_{k=a}^\infty f(k) = \sum_{k=1}^\infty f(k) - \sum_{k=1}^{a-1}f(k)$$ or more generally if $a < b < c$ $$\sum_{k=b}^c f(k) = \sum_{k=a}^c f(k) - \sum_{k=a}^{b-1}f(k)$$