Proving $(1 + 1/n)^{n+1} \gt e$

Question

I'm trying to prove that

$$ \left(1 + \frac{1}{n}\right)^{n+1} > e $$

It seems that the definition of $e$ is going to be important here but I can't work out what to do with the limit in the resulting inequality.

Answer 1

To complete The Chaz' answer:

You just need to show that the sequence $\bigl\{(1+{1\over n})^{n+1}\bigr\}$ is decreasing (one then easily shows its limit is $e$ if you know that $\bigl(1+{1\over n}\bigr)^n$ converges to $e$).

We use Bernoulli's inequality: $$ (1+x)^n>1+nx,\quad \text{for }\ \ x>-1, n\ge 1. $$

We have $$ \eqalign{

{\bigl(1+{1\over n}\bigr)^{n+1}\over \bigl(1+ {1\over n+1}\bigr)^{n+1}} &= \Bigl(1+{1\over n^2+2n}\Bigr)^{n+1}\cr & >1+{n+1\over n^2+2n}\cr & >1+{ 1\over n+1}\cr &={n+2\over n+1}. } $$ Thus $$ \Bigl(1+{1\over n+1}\Bigr)^{n+2} ={n+2\over n+1}\Bigl(1+{1\over n+1}\Bigr)^{n+1} < \Bigl(1+{1\over n}\Bigr)^{n+1}. $$ And so the sequence $\bigl{(1+{1\over n})^{n+1}\bigr}$ is decreasing.

Answer 2

Can you show that $a_n = \left ( 1 + \frac{1}{n} \right ) ^{n + 1}$ (for $n = 1, 2, 3, ...$) is a decreasing sequence that converges to $e$ ?

Then

$$\left ( 1 + \frac{1}{n} \right )^{n + 1} = \left ( 1 + \frac{1}{n} \right )^{1} \cdot\left ( 1 + \frac{1}{n} \right )^{n} $$

and taking limits (as $n \to \infty$) on both sides gives...

Answer 3

The following are lesser known facts, neverthless they are of some interest.

Let us introduce a tuning parameter $\alpha \in [0,\infty[$ and consider the sequence: $$x_\alpha (n):=\left( 1+\frac{1}{n}\right)^{n+\alpha}\; .$$ Then $\displaystyle \lim_{n\to \infty} x_\alpha (n)= e$ for any $\alpha$, but the monotonicity and the position of $x_\alpha (n)$ with respect to $e$ changes with $\alpha$ (i.e., they both can be tuned by varying $\alpha$).

Then the following statements can be proved:

1. If $1/2\leq \alpha $ then $x_\alpha (n)$ decreases strictly and converges to $e$ from above;

2. There exists a number $a\in ]0,1/2[$ s.t.:

   • if $0\leq \alpha < a$, then $x_\alpha (n)$ increases strictly and converges to $e$ from below;
   • if $a\leq \alpha <1/2$, then there exists $\nu =\nu(\alpha) \in \mathbb{N}$ s.t. $x_\alpha (n)$ decreases for $1\leq n\leq \nu$, increases for $n>\nu$ and converges to $e$ from below.

The number $a$ is something like $\ln 4 -1\approx 0.3863$. The proofs of these facts are tedious and lengthy but also elementary, for they rely on Differential Calculus.

Moreover, a simple computation with Taylor series expansion yields that $x_{1/2} (n)$ is the sequence which has the best rate of convergence to $e$ among the $x_\alpha$. In fact, we have: $$\begin{split} x_\alpha (n) -e&= \exp \left( (n+\alpha)\ \ln (1+1/n)\right) -e\\ &= \exp \left((n+\alpha) \left( \frac{1}{n}-\frac{1}{2n^2}+\text{o}(1/n^2)\right) \right) -e\\ &= \exp \left(1 +\frac{2\alpha -1}{2n} +\text{o}(1/n)\right) -e\\ &\approx \frac{e(2\alpha -1)}{2n} \end{split}$$ for $\alpha \neq 1/2$, but: $$\begin{split} x_{1/2} (n) -e&= \exp \left( (n+1/2)\ \ln (1+1/n)\right) -e\\ &= \exp \left((n+1/2) \left( \frac{1}{n}-\frac{1}{2n^2}+ \frac{1}{3n^3}+\text{o}(1/n^3)\right) \right) -e\\ &= \exp \left(1 +\frac{1}{12n^2} +\text{o}(1/n^2)\right) -e\\ &\approx \frac{e}{12n^2} \; . \end{split}$$

Answer 4

Take logarithms of both sides: $(n+1) \log(1+1/n) > 1$. With $t = 1/n$, this becomes $\log(1+t) > 1/(1+1/t) = t/(1+t)$. This is an equality at $t=0$, and the derivative of $\log(1+t) - t/(1+t)$ is $t/(1+t)^2$, which is positive for $t \ge 0$.

Answer 5

$$ e^{-1} = \left(e^{-\frac{1}{n+1}}\right)^{n+1} > \left(1-\frac{1}{n+1}\right)^{n+1} $$

and

$$ \left(1-\frac{1}{n+1}\right)^{n+1} \cdot \left(1 + \frac{1}{n}\right)^{n+1} = 1 $$

Answer 6

I would like to post a short proof of a stronger claim, namely that:

$$ \forall n\geq 1,\qquad \left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}\geq e.\tag{1} $$

By switching to logarithms, $(1)$ is equivalent to: $$ \left(n+\frac{1}{2}\right)\int_{-1/2}^{1/2}\frac{dx}{n+\frac{1}{2}+x}\,dx \geq 1 \tag{2}$$ but $g(x)=\frac{1}{x+1}$ is a convex function on $\mathbb{R}^+$, hence $(2)$ trivially follows from the Hermite-Hadamard inequality.