Question

I need help with the following sum:
sum (y^2(lambda^y*e^-lambda/y!)) from 0 to infinity

I know the answer is lambda+lambda^2. But I need to know why.

Answer 1

What's the index on the sum? y?

Answer 2

correct.

Answer 3

Wait, is it: \[
\sum_{y=0}^\infty y^2\frac{\lambda^y\cdot e^{-\lambda}}{y!}
\]

Answer 4

@wio yes. It is.

Answer 5

Hmm, well I know: \[
e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}
\]

Answer 6

Start with \[e^\lambda=\sum_{y=0}^\infty \frac{\lambda^y}{y!}\]
Differentiate this wrt lambda to get
\[\lambda e^{\lambda-1}=\sum_{y=0}^\infty y\frac{\lambda^{y-1}}{y!}\]
Multiply both sides by lambda to get
\[\lambda^2 e^{\lambda-1}=\sum_{y=0}^\infty y\frac{\lambda^y}{y!}\]

Answer 7

Do this once more to get a closed form for
\[\sum_{y=0}^\infty y^2 \frac{\lambda^y}{y!}\]
then multiple both sides by e^-lambda.

Answer 8

\[
\frac{d}{d\lambda }\lambda ^y = \lambda ^y\ln(\lambda)
\]

Answer 9

Correction: In the second line of my first post, that should be
\[e^\lambda=...\]

Answer 10

No, it's
\[\frac{d}{dy}\lambda^y=\lambda^y \ln{\lambda}\]

Answer 11

Oh, I see.

Answer 12

Okay?

Answer 13

In the end, you get
\[\lambda(\lambda+1)e^\lambda e^{-\lambda}=\sum_{y=0}^\infty y^2 \frac{\lambda^y}{y!}e^{-\lambda}\]

Answer 14

The LHS can be simplified into \(\lambda(\lambda+1)=\lambda^2+
\lambda\).

Answer 15

Thank you!