Is there a way to solve $ \sum_{n=0}^\infty \frac{x^n}{n!} $ without relying on test such as the ratio test. Possibly solve it using algebra and integrals?
Evaluating $ \sum_{n=0}^\infty \frac{x^n}{n!}$
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0Very similar question from yesterday: http://math.stackexchange.com/questions/237261/find-sum-n-1-infty-frac1n – 2012-11-15
1 Answers
As Gautam Shenoy points out in the comments, $\sum_{n=0}^\infty \frac{x^n}{n!}$ is $e^x$. One way to see this with derivatives is by noting that term-by-term differentiation gives $\frac{d}{dx}\sum_{n=0}^\infty \frac{x^n}{n!} = \sum_{n=1}^\infty \frac{nx^{n-1}}{n!} = \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!} = \sum_{n=0}^\infty \frac{x^n}{n!}$ after reindexing in the last step. In other words, it satisfies the differential equation $f' = f$. By checking with the initial condition $x=0$, the sum must be $e^x$.
(The unique solution to $f' = f$ with $f(0) = 1$ is one of the definitions of $e^x$. Another definition of $e^x$ is as the power series given. If you're using a different definition, there may be a little more work to do to use this fact.)
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1@Neal But these definitions are not equivalent. If you define $e^x$ one way, you should prove the others. – 2015-12-29