If $n$ random variables are independently distributed with an exponential distribution $f_X(x) =\lambda e^{-\lambda x} (x\geq 0)$, the range $R_n = \max(X_1,\cdots,X_n) - \min(X_1,\cdots,X_n)$ has the following expectation: $\mbox{E}[R_n] = \frac{1}{\lambda}\Big(1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n-1}\Big)$. This result is well known, see here for some context.
However, I could not find a formula for the variance, in the literature. So I decided to do the computations myself and came up with a spectacular formula. I am wondering if the following formula for the variance is original: $\mbox{Var}[R_n]= \frac{1}{\lambda^2}\Big(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots + \frac{1}{(n-1)^2}\Big)$. As a result, the limit as $n\rightarrow\infty$ is $\pi^2/(6\lambda^2)$. I have a hard time believing that this is a new result. Could someone provide a reference? I plan on including it in an upcoming article.
This result is pretty deep, though not that hard to prove. It is almost like the range, for the exponential distribution, is made up of a weighted sum of independent exponential variables with same parameter $\lambda$, each new one added into the sum contributing with a weight equal to $1/k, k=1, 2$ and so on.
