Prove that $1+x+\frac{x^2}{2}+\dots+\frac{x^n}{n!}<e^x$ for all $x\in(0,\infty),n\in\mathbb{N}$

Question

Intuitively this makes sense but I don't know how to formally show that this is true. I tried using induction but that got me nowhere .

Answer 1

$$x > 0 \implies \frac{x^k}{k!} > 0 \ \ \forall k$$ It then follows that: $$\sum_{k=0}^n \frac{x^k}{k!} < \sum_{k=0}^{n+1} \frac{x^k}{k!} < \cdots < \sum_{k=0}^\infty\frac{x^k}{k!} = e^x$$

Answer 2

This is a truncated Taylor expansion of $e^x$, and the series has all positive terms, so truncating it automatically gets you a smaller sum than the total.

Answer 3

The induction method.

$n=1$ is trivial.

Assume it holds for $n, (n\geq 1)$.

Then for $n+1$, we put $$f(x)=e^x-\left(1+x+\dots+\frac{x^{n+1}}{(n+1)!}\right)$$ By assumption, $f'(x)=e^x-(1+x+\dots+\frac{x^{n}}{n!})>0\Rightarrow f(x)>f(0)=0$ for $x>0$

Thus, it also holds for $n+1$.

Answer 4

If we officially do not yet know about Taylor series, we can use induction. First we prove the base case, that $1\lt e^x$ for all $x\gt 0$. Let $f(x)=e^x-1$. Then $f(0)=0$, and $f'(x)=e^x\gt 0$ for all $x$. So $f(x)$ is increasing on our interval, and therefore $f(x)\gt 0$ for all positive $x$.

Now suppose that we know that for a particular $k$ we have $$1+x+\frac{x^2}{2!}+\cdots +\frac{x^k}{k!}\lt e^x\tag{1}$$ for all positive $x$. We want to show that $$1+x+\frac{x^2}{2!}+\cdots +\frac{x^{k+1}}{(k+1)!}\lt e^x\tag{2}$$ for all positive $x$. Let $$f(x)=e^x-\left(1+x+\frac{x^2}{2!}+\cdots +\frac{x^{k+1}}{(k+1)!}\right).$$ Note that $f(0)=0$, and that $$f'(x)=e^x-\left(1+x+\frac{x^2}{2!}+\cdots +\frac{x^{k}}{k!}\right).$$ By (1), $f'(x)\gt 0$ for all $x$, so $f$ is increasing. This shows that for $x\gt 0$ the inequality (2) holds.

Answer 5

This is easily solved once recognizing the finite sum as a partial sum of the series that defines $e^x$