Need help calculating $\sum_{i=0}^\infty \dfrac{e^{-\lambda}\lambda^{2i}}{(2i)!}, 0 \lt\lambda \lt 1$

I am trying to calculate the following sum:

$$\sum_{i=0}^\infty \dfrac{e^{-\lambda}\lambda^{2i}}{(2i)!}, 0 \lt\lambda \lt 1$$

It is clearly a power series, however I am stuck trying to move beyond this conclusion. I tried to solve it using wolfram alpha and it gives the following answer: $e^{-\lambda} \cosh(\lambda)$. What would be the approach to get the answer the wolfram alpha gives me?

Thanks!

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2 Answers

Factor out $e^{-\lambda}$ and notice that you have even series from a Taylor expansion, which is equivalent to $0.5*(e^x + e^{-x})$.

Edit To make this more detailed, notice that in the Taylor expansion

$$e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$

you would like to keep even terms and remove the odd terms. Note that

$$e^{-x} = \sum_{k=0}^\infty \frac{(-1)^k x^k}{k!} = \sum_{k \text{even}}\frac{x^k}{k!} - \sum_{k \text{ odd}}\frac{x^k}{k!},$$

since $(-1)^k$ is $1$ for even $k$ and $-1$ for odd $k$. Now you can add the two series, and the odd terms will cancel, exactly as you need. The problem is, you will have each even twice, not once. To deal with that, divide by $2$, getting

$$\frac{e^x + e^{-x}}{2} = \sum_{k \text{even}}\frac{x^k}{k!} = \sum_{i = 0}^\infty \frac{x^(2i)}{(2i)!},$$

as desired, since the left-hand side is exactly $\cosh x$.

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Ok, I can see that it is the Taylor series expansion now. But how exactly do I move to the 0.5 * (e^x+e^-x) conclusion? Thanks! – bim_bam Sep 7 '12 at 23:04
@bim_bam: i made an edit, making the answer more detailed. – gt6989b Sep 9 '12 at 20:12

Note that the term $e^{-\lambda}$ doesn't have any dependence on $i$, so you can move it out front of the sum.

Now write down the power series for $cosh(\lambda)$.

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