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It is well known that we can define $e^x$ by the following limit

$$e^{x}=\lim_{n\to\infty}\left(1+{x\over n}\right)^n$$

I would like to show that the RHS sequence is always less than or equal to $e^x$ for all $-n\le x\le0$ and $n>1$, and what I had currently done is to study the property of this sequence (which is defined as $f(x)$)

$$f(x)=\left(1+{x\over n}\right)^n$$

I also find its first and second derivative $f'(x)=\left(1+{x\over n}\right)^{n-1}$ and $f''(x)={n-1\over n}\left(1+{x\over n}\right)^n$ and show that they are strictly positive for all $x\in[-n,0]$. As a result $f(x)$ is monotonically increasing and concave up. By plugging in the end points I found that $f(0)=1\le e^0$ and $f(-n)=0<e^{-n}$, I wonder if these conditions allow me conclude that $f(x)\le e^x$ for all $x\in[-n,0]$.

TravorLZH
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3 Answers3

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To show that $(1+x/n)^n\leq e^x$ for all $x\geq -n$ is equivalent to showing that $$ 1+\frac{x}{n}\leq e^{x/n},\qquad x\geq -n. $$ This is a special case of the following inequality $$ 1+y\leq e^y,\qquad (\star) $$ which holds for all $y\in\mathbb R$, when we substitute $y=x/n$.

To prove $(\star)$, you can observe that it is an equality when $y=0$, and the derivative of the left side is less than the derivative of the right side for all $y>0$, and it is larger for $y<0$.

pre-kidney
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Here's an elementary proof that only requires Bernoulli's inequality.

Let $\displaystyle u_n:=\left(1+\frac xn\right)^n$ and let $N$ be the smallest integer strictly greater than $|x|$. Consider $n\geq N$.

Then $n>|x|$, hence $0<1+\frac xn$. Besides, $$u_{n+1}=\left(1+\frac xn \right)^{n+1}\left(1-\frac{x}{n(n+1)\left(1+\frac xn\right)} \right)^{n+1}$$ Note that for $n\geq N$, $$-\frac{x}{n(n+1)\left(1+\frac xn\right)}\geq -1 \iff \frac{x}{n(n+1)} \leq 1+\frac xn$$ and $$\frac{x}{n(n+1)} \leq 1+\frac xn \iff -x\leq n+1$$

Bernoulli's inequality applies and yields $\forall n\geq N$, $$\left(1-\frac{x}{n(n+1)\left(1+\frac xn\right)} \right)^{n+1}\geq 1-\frac{x}{n\left(1+\frac xn\right)}$$ Since $\displaystyle 1-\frac{x}{n\left(1+\frac xn\right)} = \frac{1}{1+\frac xn}$, we get that for $n\geq N$, $$u_{n+1}\geq u_n$$

Gabriel Romon
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Let $x=-y$ so that $0\leq y\leq n$. If $a_n$ is the sequence in question then consider $$b_n=\frac{1}{a_n}=\left(1-\frac{y}{n}\right)^{-n}=1+y+\frac{1/n+1}{2!}y^2+\frac {(1/n+1)(1/n+2)}{3!}y^3+\dots$$ via binomial theorem. Clearly we can see that $b_n$ is decreasing and hence $a_n$ is increasing and therefore does not exceed its limit $e^x$.

Paramanand Singh
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