Prove $e^x$ is its own derivative via power series?

Question

$$ \frac{d}{dx}e^x =\frac{d}{dx} \sum_{n=0}^{ \infty} \frac{x^n}{n!}$$ $$ \sum_{n=0}^{ \infty} \frac{nx^{n-1}}{n!}$$ $$ \sum_{n=0}^{ \infty} \frac{x^{n-1}}{(n-1)!}$$ This isn't as straightforward as I thought it would be. I'm assuming there is some way to shift indices to complete the proof, but I'm not sure exactly how.

Answer 1

Well, first of all, careful with your first term, since you are dividing by $(-1)!$. The first term is constant and differentiates to $0$, so your sum should start from $n = 1$.

Then, just make a substitution $m = n - 1$. When $n = 1$, we get $m = 0$, and so

$$\frac{\mathrm{d}}{\mathrm{d}x} e^x = \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!} = \sum_{m=0}^\infty \frac{x^m}{m!}.$$

Answer 2

Rewrite the infinite sum as $$\sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + \sum_{n=1}^{\infty}\frac{x^n}{n!}$$ Then $$\frac{d}{dx} (1 + \sum_{n=1}^{\infty}\frac{x^n}{n!}) = \sum_{n=1}^{\infty}\frac{nx^{n-1}}{n!} = \sum_{n=1}^{\infty}\frac{x^{n-1}}{(n-1)!}=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$

Answer 3

Seems to me people are giving correct versions of the calculation, hence showing that your conclusion is wrong, but without pinning down exactly where the error in your version is. The error is here: The "identity" $$\frac k{k!}=\frac1{(k-1)!}$$is not true if $k=0$. (You obtain that identity by cancelling a $k$ in the numerator and denominator. Ignoring the fact that if $k=0$ there is no factor of $k$ in $k!$, in general you cannot "cancel" zeroes: $1(0)=2(0)$, hence $1=2$.)