# Prove $\left(1+\frac{x}{n}\right)^n < e^x$, where $x$ is any positive real number and $n$ is any positive integer.

I am having trouble with my homework problem, it says:

Suppose that $n$ is a positive integer and that $x > 0$. Show that $$\left(1+\frac{x}{n}\right)^n < e^x.$$

I have proved the base case, but I am unsure how to proceed with the induction step. The question continues:

and, if $x < n$, then $$e^x < \left(1-\frac{x}{n}\right)^{-n}$$ Also, by choosing a suitable $n$, deduce that $2.5 < e < 2.99$.

I would like a hint to help me along if anyone can help. Thanks!

• Which definition of $e$ are you allowed to use? – angryavian Feb 7 '16 at 22:45

For the first one, $\left(1+\frac{x}{n}\right)^n < e^x \iff 1+\frac{x}{n} < e^{x/n}$ and this follows from $e^x \ge 1+x$ with equality only if $x = 0$.

For the second one, $e^x < \left(1-\frac{x}{n}\right)^{-n} \iff e^{x/n} < \frac1{1-\frac{x}{n}}$ and this follows from $e^x < \frac1{1-x}$ if $0 < x < 1$. This, in turn, follows from $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ and $\frac1{1-x} = \sum_{n=0}^{\infty} x^n$ if $0 < x < 1$.

• For an arguably more symmetric approach, $e^x<\frac{1}{1-x}$ can also be established by noting its equivalence to $1-x<e^{-x}$, which is simply $e^x>1+x$ with $x\to-x$. – πr8 Feb 13 '16 at 10:50

Just see this for $x>0$

$$e^{n\ln(1+x/n)} < e^{n(x/n)} =e^x$$

since $\ln(1+t)<t$.

Bernoulli's Inequality implies that $$\left(1+\frac xn\right)^n$$ is increasing in $n$ for $x\gt-n$. This implies that for $x\gt0$ $$\left(1+\frac xn\right)^n\le e^x$$ and for $n\gt x\gt0$ $$\left(1-\frac xn\right)^n\le e^{-x}$$ and so $$\left(1-\frac xn\right)^{-n}\ge e^x$$