Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to calculate this sum:


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?

share|cite|improve this question
up vote 8 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.

share|cite|improve this answer
Thank you very much! Helped me a lot! – Aufwind Dec 11 '11 at 23:53

$$1+\sum_{k=1}^\infty\frac{\lambda^k}{k!}=\sum_{k=0}^\infty\frac{\lambda^k}{k!}=\mathrm e^\lambda. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.