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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.

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What definition of $e$ are you using? – Aryabhata Mar 16 '12 at 21:14
$lim_{n \rightarrow \inf} (1 + 1/n)^n$ – maxaposteriori Mar 16 '12 at 21:16
From a back-of-the-envelope calculation: your sequence is strictly monotonously falling (consider the difference between adjacent sequence elements), and its limits is $e$. Hence, each element of the sequence must be larger than $e$. – Johannes Kloos Mar 16 '12 at 21:27
As it's the only non-trivial part of the argument, I think the "envelope" should be shown. – David Mitra Mar 16 '12 at 21:43
up vote 12 down vote accepted

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 $e$ ?


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

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

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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.

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(+1) This was the part that was "left to the reader" of my envelope... ; ) – The Chaz 2.0 Mar 16 '12 at 22:11
Thanks for the elaboration--this was helpful. – maxaposteriori Mar 17 '12 at 14:06
This may sound like a stupid question but how did you do this: ${\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} $? – GinKin Nov 17 '13 at 21:04
@GinKin Use $ {a^x\over b^x} =(a/b)^x$ and$$ {1+{1\over n}\over 1+{1\over n+1}} ={{n+1\over n}\over {n+1+1\over n+1}}={(n+1)^2\over n(n+2)} ={ (n^2+2n) +1\over n(n+2)} =1+{1\over n^2+2n}. $$ – David Mitra Nov 17 '13 at 21:11
Um shouldn't it be $\eqalign{ {\bigl(1+{1\over n}\bigr)^{n+1}\over \bigl(1+ {1\over n+1}\bigr)^{n+\color{red}2}} }$? – GinKin Mar 23 '14 at 11:01

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 $t \ge 0$.

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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: $$\begin{split} x_{1/2} (n) -e&= \exp \left( (n+1/2)\ \ln (1+1/n)\right) -e\\ &= \exp \left((n+1/2) \left( \frac{1}{n}-\frac{1}{2n^2}+ \frac{1}{3n^3}+\text{o}(1/n^3)\right) \right) -e\\ &= \exp \left(1 +\frac{1}{12n^2} +\text{o}(1/n^2)\right) -e\\ &\approx \frac{e}{12n^2} \; . \end{split}$$

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


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

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It's not clear to me why your first inequality is true. – Patrick Mar 16 '12 at 23:20
@Patrick $e^x \geq 1+x$ with equality only for $x=0$. – WimC Mar 17 '12 at 12:14
Got it. That's what I was missing. – Patrick Mar 17 '12 at 20:06
@WimC how can we just assume that $e^x \geq 1+x$? – Dmitri Nesteruk Aug 31 '14 at 15:35
@DmitriNesteruk Not just assume. It depends a bit on how you introduce the exponential function. I like to see it as the inverse of $\log$ and then show $\log(x) \leq x-1$ first. If you define it as the limit of $(1+\tfrac{x}{n})^n$ then it is Bernoulli's inequality. – WimC Aug 31 '14 at 18:44

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.

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