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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.

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  • $\begingroup$ Why would you not be able to use the geometric series formula? The variables / constants in the exponent should be positive (in a physical context, at least) so this should fall within the radius of convergence. Although, I don't see where the $n$ and $k$ end up in the formula... is the $i$ index supposed to be $k$, and $n$ be $\infty$? $\endgroup$ – Tob Ernack Jun 6 '17 at 15:30
  • $\begingroup$ Presumably this is an infinite series. You can view this as a geometric series by rewriting your summand as $$\left(e^{-h\nu/T}\right)^k.$$ $\endgroup$ – Cameron Williams Jun 6 '17 at 15:31
  • $\begingroup$ I have taken the liberty of changing the letter $v$ to the letter $\nu,$ since that is a somewhat standard usage in this context. $\endgroup$ – Michael Hardy Jun 6 '17 at 15:32
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\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}

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    $\begingroup$ Thanks for that. For some reason I thought I couldnt use that formula. Also nice to see where the standard formula for the sum of a finite geometric series actually comes from. $\endgroup$ – goml Jun 6 '17 at 15:51
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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.

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  • $\begingroup$ note where $|x|<1$ $\endgroup$ – Dando18 Jun 6 '17 at 15:33

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