The Notes: `sum` (2^n)/n!

Friday, November 13, 2009

$\displaystyle\sum$ (2^n)/n!

Hello!

I assume the sum is from $\displaystyle{n}={0}$ to infinity. This series is (absolutely) convergent, which is simple to check by the Ratio test.

Consider the exponential function $\displaystyle{{e}}^{{x}}.$ Its Taylor series is $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{x}}^{{n}}}{{{n}!}},$ and $\displaystyle{{e}}^{{x}}$ is equal to the sum of this series for all real (and complex) $\displaystyle{x}.$ The sum in question is therefore $\displaystyle{{e}}^{{2}},$ which is approximately $\displaystyle{7.39}.$

The Notes

Friday, November 13, 2009

$\displaystyle\sum$ (2^n)/n!

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