This problem involves evaluating a sum that appears in the context of a Poisson distribution. The given sum is:

$\sum_{k=0}^\infty k \frac{\lambda^k e^{-\lambda}}{k!}.$

Steps to Solve:

1. Factor the terms: $k \frac{\lambda^k e^{-\lambda}}{k!} = \frac{\lambda^k e^{-\lambda}}{(k-1)!}.$

   For $k = 0$, the term is 0, so the sum starts effectively from $k = 1$.

   The sum can now be rewritten as: $\sum_{k=1}^\infty \frac{\lambda^k e^{-\lambda}}{(k-1)!}.$

2. Substitute $m = k-1$: Let $m = k-1$, so when $k = 1$, $m = 0$. The sum becomes: $\sum_{m=0}^\infty \frac{\lambda^{m+1} e^{-\lambda}}{m!}.$

3. Factor out $\lambda$: $\lambda e^{-\lambda} \sum_{m=0}^\infty \frac{\lambda^m}{m!}.$

4. Recognize the series: The term $\sum_{m=0}^\infty \frac{\lambda^m}{m!}$ is the Taylor expansion of $e^\lambda$. Thus: $\sum_{m=0}^\infty \frac{\lambda^m}{m!} = e^\lambda.$

5. Simplify the result: Substituting this back: $\lambda e^{-\lambda} \cdot e^\lambda = \lambda.$

Final Answer:

$\sum_{k=0}^\infty k \frac{\lambda^k e^{-\lambda}}{k!} = \lambda.$

Let me know if you'd like a more detailed breakdown or have questions! Here are some follow-up questions for practice:

1. How would the result change if the summation started from $k = 2$?
2. Can you derive this result using differentiation of a generating function?
3. What is the relevance of this sum in the context of Poisson distribution mean?
4. What does this result tell us about the expected value of $k$ in this distribution?
5. How would the computation differ if $k^2$ appeared in the numerator instead of $k$?

Tip: Recognizing Taylor series expansions like $e^\lambda$ is crucial in simplifying sums involving exponential and factorial terms!