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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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    $\begingroup$ What definition of $e$ are you using? $\endgroup$
    – Aryabhata
    Mar 16, 2012 at 21:14
  • $\begingroup$ $lim_{n \rightarrow \inf} (1 + 1/n)^n$ $\endgroup$ Mar 16, 2012 at 21:16
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    $\begingroup$ 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 $e$. $\endgroup$ Mar 16, 2012 at 21:27
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    $\begingroup$ As it's the only non-trivial part of the argument, I think the "envelope" should be shown. $\endgroup$ Mar 16, 2012 at 21:43

6 Answers 6

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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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  • $\begingroup$ (+1) This was the part that was "left to the reader" of my envelope... ; ) $\endgroup$ Mar 16, 2012 at 22:11
  • $\begingroup$ Thanks for the elaboration--this was helpful. $\endgroup$ Mar 17, 2012 at 14:06
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    $\begingroup$ 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} $? $\endgroup$
    – GinKin
    Nov 17, 2013 at 21:04
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    $\begingroup$ @GinKin Use $ {a^x\over b^x} =(a/b)^x$ and$$ {1+{1\over n}\over 1+{1\over n+1}} ={{n+1\over n}\over {n+1+1\over n+1}}={(n+1)^2\over n(n+2)} ={ (n^2+2n) +1\over n(n+2)} =1+{1\over n^2+2n}. $$ $\endgroup$ Nov 17, 2013 at 21:11
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    $\begingroup$ @DavidMitra This use of Bernoulli's inequality seems to be standard textbook fare, but is there an actual attribution for this approach to the proof? Bernoulli did not use this approach, right? $\endgroup$
    – Alan
    Jun 30, 2018 at 18:45
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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...

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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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    $\begingroup$ Can you give me the proof thai this function is stricly decreasing when $\alpha = 1/2$? Thank you in advance. $\endgroup$
    – Mr.Lilly
    Sep 12, 2016 at 16:24
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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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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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  • $\begingroup$ Nice use of the HHI. (+1) $\endgroup$
    – Mark Viola
    Feb 27, 2017 at 20:33
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$$ e^{-1} = \left(e^{-\frac{1}{n+1}}\right)^{n+1} > \left(1-\frac{1}{n+1}\right)^{n+1} $$

and

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

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  • $\begingroup$ It's not clear to me why your first inequality is true. $\endgroup$
    – Patrick
    Mar 16, 2012 at 23:20
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    $\begingroup$ @Patrick $e^x \geq 1+x$ with equality only for $x=0$. $\endgroup$
    – WimC
    Mar 17, 2012 at 12:14
  • $\begingroup$ Got it. That's what I was missing. $\endgroup$
    – Patrick
    Mar 17, 2012 at 20:06
  • $\begingroup$ @WimC how can we just assume that $e^x \geq 1+x$? $\endgroup$ Aug 31, 2014 at 15:35
  • $\begingroup$ @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. $\endgroup$
    – WimC
    Aug 31, 2014 at 18:44

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