$\sum_1^\infty\frac{(k\theta e^{-\theta})^k}{k!}=\frac{\theta}{1-\theta}$, where $0<\theta<1$.
It can be verified via simulation, but I haven't proved it.
Are there any previous results on this infinite sum?
Thanks!
$\begingroup$
$\endgroup$
7
-
1$\begingroup$ For what range of values of $\theta$? For which $\theta$ does the series converge? $\endgroup$– Gerry MyersonSep 9, 2013 at 9:35
-
$\begingroup$ Yes, it should be $0<\theta<1$ $\endgroup$– Severals-user45972Sep 9, 2013 at 9:49
-
$\begingroup$ Putting $x = \theta e^{-\theta}$, the left hand side is very close to the Lambert W function en.wikipedia.org/wiki/Lambert_W_function . $\endgroup$– David E SpeyerSep 9, 2013 at 18:22
-
$\begingroup$ Thanks, its done. The steps are as follows,$\sum_1^\infty \frac{(-kx)^k}{k}=-x\sum_1^\infty \frac{(-k)^{k-1}}{k} k x^{k-1}=-x W_0'(x)=-x\frac{W_0(x)}{x(1+W_0(x))}$. Note that $W(x)=W(-\theta e^{-\theta})=-\theta$. $\endgroup$– Severals-user45972Sep 10, 2013 at 1:50
-
$\begingroup$ Wait a minute, there's a factorial in the denominator in the question, but just $k$ in your comment. What gives? $\endgroup$– Gerry MyersonSep 10, 2013 at 13:21
|
Show 2 more comments