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I am learning about Poisson Random Variable and came across a problem with this infinite sum:

$\sum^{∞}_{n=2}\frac{e^{-2}(2)^i}{i!}$

The first step I do is move the constant $e^{-2}$ out.

$e^{-2} * (\sum^{∞}_{n=2}\frac{(2)^i}{i!})$

But I don't know what is the next step after this, I am trying to use this formula: $\frac{a}{1-r}$

I looked through my notes from class, but I can only find basic example that cover $\sum^{∞}_{n=1} 2x$. In the this case, what would be the ratio?

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  • $\begingroup$ Hint: What is the Taylor series for the exponential function? $\endgroup$ – Robert Israel Mar 10 '16 at 23:50
  • $\begingroup$ HINT: $$\sum_{i=0}^{\infty} \frac{x^i}{i!} = e^x$$ $\endgroup$ – Crostul Mar 10 '16 at 23:51
  • $\begingroup$ Shouldn't the $n$ in the index of the sum be an $i$? $\endgroup$ – Ryan Goulden Mar 30 '17 at 6:18
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Using the series expansion for the exponential function, $$e^{x} = \sum_{r=0}^{\infty} \frac{x^{r}}{r!},$$ then by subtracting the first two terms from both sides: $$\sum_{r=2}^{\infty} \frac{x^{r}}{r!} = e^{x} - 1 - x.$$

Now, \begin{align} \sum_{n=2}^{\infty} \frac{e^{-2} \, 2^{n}}{n!} &= e^{-2} \, \left(e^{2} - 1 - 2 \right) \\ &= 1 - 3 \, e^{-2}. \end{align}

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It will certainly help to know that $$ e^x=\sum_{i=0}^\infty \frac{x^i}{i!}$$

Just fill in $x=2$ and work from there!

The equality I use is called the taylor series at $0$ or the maclaurin series of $e^x$. It's not hard to derive.

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