Find the sum for the following series $\sum_{k=0}^n e^{-hk\nu/T}$ Find the sum for the following series:
$$\sum_{k=0}^\infty e^{-hk\nu/T} $$
where:
$h =$ Planck's constant 
$\nu =$ the frequency of the oscillator
$T =$ temperature in kelvin.
Answer: 
$$\frac{1}{1-e^{-h\nu/T}} $$
The problem is that I have no idea how they got the answer.
I guess I am confused because of the extra variables. I assumed that I can't use the formula for a geometric series and in previous exercises I had to write the series as a telescoping series but I am not sure if this is possible in this case.
Any help would be appreciated.
 A: For $|x|<1$ we know for the geometric series that: $$\sum_{k=0}^{+\infty}x^k = \frac{1}{1-x}$$
Now you can use the substitution $x = e^{-h\nu/T}$ to obtain your result.
A: \begin{align}
\sum_{k=0}^n e^{-hk\nu/T} & = \sum_{k=0}^n r^k \quad \text{where } r = e^{-h\nu/T} \\[10pt]
& = \frac{1-r^{n+1}}{1-r} \tag 1 \\[10pt]
& \to \frac 1 {1-r} \text{ as } n\to \infty, \text{ if } |r|<1 \\[10pt]
& = \frac 1 {1 - e^{-h\nu/T}}
\end{align}
The equality $(1)$ is the standard formula for the sum of a finite geometric series, and can be proved by observing the cancelation of all except the first term and the last in the expression that results from multiplying both sides of the equality by $1-r.$
The inequality $|r|<1$ holds because $h\nu/T>0.$
Here's a way to see that $1+r+r^2+r^3+\cdots+r^n = \dfrac{1-r^{n+1}}{1-r}:$
\begin{align}
& (1+r+r^2+r^3+\cdots+r^n)(1-r) \\[10pt]
= {} & \Big(1+r+r^2+r^3+\cdots+ r^n\Big)\cdot 1 - \Big( 1 + r + r^2 + r^3 + \cdots + r^n \Big)\cdot r \\[10pt]
= {} & \Big(1+r+r^2+r^3+\cdots+ r^n\Big) - \Big( r+r^2+r^3 + r^4 + \cdots + r^n + r^{n+1} \Big) \\[10pt]
= {} & 
1 - r^{n+1} \text{ because all terms cancel except the first and the last.}
\end{align}
