Solving $$\sum_{k=0}^\infty \frac{k(vt)^ke^{-vt}}{k!}$$ where v is a constant. How is the answer equals vt?
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Hint:$$e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$
M.H
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Your answer seems to be incorrect I DO NOT understand how you can work on variable $k$ outside the sign of summation. See my edit and my answer. – Adi Dani Dec 12 '13 at 13:13
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I've been looking at it since that It is not possible to work on variable k outside its sign of summation. – user107676 Dec 12 '13 at 16:26
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$$\sum_{k=0}^\infty \frac{k(vt)^ke^{-vt}}{k!}=\sum_{k=1}^\infty \frac{k(vt)^ke^{-vt}}{k!}$$ $$=vte^{-vt}\sum_{k=1}^\infty \frac{(vt)^{k-1}}{(k-1)!}=vte^{-vt}\sum_{j=0}^\infty \frac{(vt)^{j}}{j!}=$$ $$=vte^{-vt}e^{vt}=vt$$
Adi Dani
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