# Evaluating $\sum_{n=0}^\infty \frac{x^n}{n!}$

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?

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Thats $e^x$ expansion. – Gautam Shenoy Nov 15 '12 at 6:00
As Gautam Shenoy says, it is $e^x$ but if you are asking this we need your definition of $e^x$. There are a number of routes that lead the same place. – Ross Millikan Nov 15 '12 at 6:04
The ratio test only tells you that it converges, not what it converges to, so if you want a value you need something else. – Ross Millikan Nov 15 '12 at 6:07
– Ragnar Nov 15 '12 at 6:12
Very similar question from yesterday: math.stackexchange.com/questions/237261/… – Hans Lundmark Nov 15 '12 at 6:29

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.)
Even the standard functions have to be defined! You can define $e^x$ as the unique function satisfying $f'=f, f(0)=1$, you can define $e^x = \sum \frac{x^n}{n!}$, you can define $e^x = \lim_{n\to\infty} (1 + \frac{x}{n})^n$, you can define $e^x$ as the number $y$ such that $\int_1^y\frac{du}{u} = x$, ... see en.wikipedia.org/wiki/… – Neal Nov 15 '12 at 6:21
@Neal But these definitions are not equivalent. If you define $e^x$ one way, you should prove the others. – Nima Bavari Dec 29 '15 at 19:35