In the proof by Denis Hanson (see Lemma 3), this is presented as obvious if $a \ge 2$ and $n > a$:

$$\left(1 + \dfrac{1}{(n-a+1)/(a-1)}\right)^{(n-a+1)/(a-1)} < e$$

where $e$ is Euler's constant which is approximately $2.72$

For me, this is not obvious so I thought I would detail the argument for why this is true.

Here's what I came up with (I understand that induction can be used for real numbers under certain circumstances):

  • Let $u = (n-a+1)/(a-1)$

  • Base Case: for $n=a$, then $u = \dfrac{1}{a-1}$

$\left(1 + \dfrac{1}{u}\right)^u = a^{\frac{1}{a-1}} < 2 < e$ since $\dfrac{\log a}{\log 2} \le a-1 $

  • Inductive Hypothesis: up to some $u \ge \dfrac{1}{a-1}$: $$\left(1 + \dfrac{1}{u}\right)^{u} < e$$

  • Inductive case: For $\epsilon > 0$:

$\dfrac{u+1}{u} > \dfrac{u+1+\epsilon}{u+\epsilon}$ since $(u+1)(u+\epsilon) = u^2 + u + u\epsilon + \epsilon > u^2 + u + u\epsilon = (u)(u+1+\epsilon)$

$\left(\dfrac{u+1}{u}\right)^u > \left(\dfrac{u+\epsilon+1}{u+\epsilon}\right)^u$

$\left(\dfrac{u+\epsilon+1}{u+\epsilon}\right)^{\epsilon} < 1$

Which gives us that:

$$\left(1 + \dfrac{1}{u+\epsilon}\right)^{u+\epsilon}< \left(1+\dfrac{1}{u}\right)^{u} < e$$

Is this argument valid? Is there a simpler way to establish Hanson's step?

  • What denotes $e$ here? – Dr. Sonnhard Graubner Oct 5 at 17:06
  • $e$ is Euler’s constant ~ 2.72.... I’ll update the question to add this detail. – Larry Freeman Oct 5 at 17:13
  • 1
    Put $t = (a-1)/(n+ a -1) \in (0,1)$. The estimate is equivalent to $(1+t)^{1/t} < e$ or $\ln(1+t) < t$. The function $f(t) = \ln(1+t) -t$ satisfies $f(0) = 0$ and $f'(t) = -t/(1+t) < 0$. Therefore $f(t) < f(0) =0$ for $t >0$. – nguyen0610 Oct 5 at 17:28
  • Thanks! If you put that as the answer, I will be glad to accept. – Larry Freeman Oct 5 at 17:39
  • In this answer it is shown that $\left(1+\frac1n\right)^n$ is increasing and $\left(1+\frac1n\right)^{n+1}$ is decreasing. – robjohn Oct 5 at 18:05
up vote 3 down vote accepted

Put $t = (a-1)/(n+a -1) \in (0,1)$. The estimate is equivalent to $(1+ t)^{1/t} < e$ or $\ln(1+t) < t$. Consider the function $f(t) = \ln(1+t) -t, t >0$. We have $f(0) =0$ and $$f'(t) = \frac1{1+t} -1 = -\frac{t}{t+1} < 0, \, t >0.$$ Therefore, $f(t) < f(0) =0$ for $t>0$, or equivalently, $\ln(1+t) < t$ for $t >0$.

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