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I am trying to calculate this sum:

$$\sum_{k=1}^\infty\frac{q\lambda^k}{k!e^\lambda}$$

My solution is:

$$\begin{align*} \sum_{k=1}^\infty\frac{q\lambda^k}{k!e^\lambda}&=\frac{q}{e^\lambda}\sum_{k=1}^\infty\frac{\lambda^k}{k!}=\frac{q}{e^\lambda}\sum_{k=0}^\infty\frac{\lambda^{k+1}}{(k+1)!}\\ &=\frac{q\lambda}{e^\lambda\cdot (k+1)}\sum_{k=0}^\infty\frac{\lambda^k}{k!}\lambda=\frac{q\lambda}{e^\lambda\cdot (k+1)}\cdot e^\lambda=\frac{q\lambda}{k+1}\end{align*}$$

But the solution given by my tutor is $q(1-e^{-\lambda})$. Could someone please verify my calculation and tell me where I went wrong?

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2 Answers

up vote 7 down vote accepted

You're correct as far as

$$ {q \over e^\lambda} \sum_{k=0}^\infty {\lambda^{k+1} \over (k+1)!} $$

but then you pull $k+1$ out of the sum. Since $k+1$ is not a constant -- you're summing over $k$ -- you can't do that.

I'd go back to

$$ {q \over e^\lambda} \sum_{k=1}^\infty {\lambda^k \over k!} $$

and then notice (as you have!) that the sum here is similar to the Taylor series for $e^z$ evaluated at $z = \lambda$. In fact, we can rewrite this as

$$ {q \over e^\lambda} \left( \left( \sum_{k=0}^\infty {\lambda^k \over k!} \right) - {\lambda^0 \over 0!} \right) $$

and the sum here is now $e^\lambda$. So the original sum is

$$ {q \over e^\lambda} \left( e^\lambda - 1 \right) $$

which can be simplified to the answer given by your tutor.

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Thank you very much! Helped me a lot! –  Aufwind Dec 11 '11 at 23:53
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$$1+\sum_{k=1}^\infty\frac{\lambda^k}{k!}=\sum_{k=0}^\infty\frac{\lambda^k}{k!}=\mathrm e^\lambda. $$

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