# Deriving the Taylor Series for $e^x$ by using integration

I am playing around with using elementary techniques to derive the Taylor Series for $$e^x$$.

Consider the sequence of integrals $$I_n = \int_0^x t^n e^{-t} dt$$ It can be shown by induction that $$I_n = n! \left ( 1-\frac{1}{e^x}\sum_{k=0}^{n} \frac{x^k}{k!}\right)$$

I want to then consider taking $$n\to \infty$$ to establish the Taylor Series. This of course relies on $$\lim_{n\to\infty} \frac{I_n(x)}{n!}=0$$ which doesn't seem to easy to show for all $$x$$.

Any ideas would be fantastic! I am hoping that there is something elementary to use here.

• An easier way would just be to consider $D(e^t) = e^t$ and just see what that does to the power series coefficient-by-coefficient and you probably need boundary condition that $e^0=1$ to get the dominoes rolling. May 12 '20 at 11:24
• You might want to take a look at the Gamma function, using $\int_0^\infty t^n\exp(-t)dt=\Gamma(n+1)=n!$ May 12 '20 at 11:25
• Curious method. For $x$ positive $e^{-t}\leq 1$ and you get a very good estimate. For $x$ negative, on the other hand, your method does not give good error estimates (and dividing by $e^x$ is probably not a good idea in that case). May 12 '20 at 11:27
• For positive $x$, you can show that $\int_0^x t^n e^{-t} dt \lt \int_0^{\infty} t^n e^{-t} dt = n!$ May 12 '20 at 11:30

$$\left|\frac{I_n}{n!}\right | = \left| \int_0^x \frac{t^n}{n!}e^{-t}\:dt\right | \leq \frac{|x|^{n+1}}{n!} \to 0$$
• @AnInvisibleCarrot $$\int_a^b f(x)\:dx \leq \sup_{x\in[a,b]} f(x) \cdot (b-a)$$ May 12 '20 at 11:41
• So wouldn't $|\frac{I_n}{n!}| \leq \frac{ |x|^{n+1} }{n!} \times \max \{ 1 , e^{-x} \} \leq \frac{ |x|^{n+1} e^{|x|}}{n!}$. Your logic does follow though with what I need to do - thank you! May 12 '20 at 12:28