# The maximum value of $p>0$ such that $(1+\frac{1}{n})^{n+p}<e$

Find the maximum value of $$p>0$$ such that $$(1+\frac{1}{n})^{n+p} for all positive integer.

I don’t really know how to start this problem. I think an idea is to consider some function like $$f(x) = \log_{1+\frac{1}{n}}{\frac{e}{(1+\frac{1}{n})^n}}$$, $$f(n)>p$$, analyze it using derivatives and find its minimum, but I’m not sure. There might be some other ways. Can you help me?

• What do you mean with "all n positive integer"?
– user
Mar 15, 2021 at 13:55
• Every integer $n>0$ satisfies that inequality for a fixed $p$ and we need the maximal $p$ so that the property given is true for every such $n$. Mar 15, 2021 at 14:11
• What do you mean by "every such 𝑛"?
– user
Mar 15, 2021 at 14:16
• I meant that every positive integer n must satisfy that inequality for a fixed parameter p. Mar 15, 2021 at 14:34
• Could you explain a bit more? If we make the limit of both sides, it might be possible, both being e. Mar 15, 2021 at 14:54

As will be shown below no such $$p$$ exists. However we indeed can find such $$p>0$$ that the inequality $$\left(1+\frac1n\right)^{n+p}\color{red}\le e$$ is valid for any postive integer.

Preliminary we prove the following

Lemma 1:

$$\forall x\ge0:\quad\frac{x}{\sqrt{1+x}}\ge \log(1+x).\tag1$$

Proof:

Let $$u(x)=\dfrac{x}{\sqrt{1+x}}-\log(1+x)$$. Obviously $$u(0)=0$$ and $$u'(x)=\frac1{(1+x)^{1/2}}-\frac x{2(1+x)^{3/2}}-\frac1{1+x} =\frac{2+x-2\sqrt{1+x}}{2(1+x)^{3/2}}=\frac{(\sqrt{1+x}-1)^2}{2(1+x)^{3/2}}\ge0,$$ so that claim $$(1)$$ is proved. Observe that $$(1)$$ is a strict inequality for all $$x>0$$.

Substituing in $$(1)$$ $$t=\dfrac1x$$ one obtains equivalent form: $$\forall t>0:\quad\frac{1}{\sqrt{t(1+t)}}> \log\left(1+\frac1t\right) \iff \sqrt{t(1+t)}\,\log\left(1+\frac1t\right)<1.\tag2$$

Consider now the following equivalent inequalities: $$\left(1+\frac1n\right)^{n+p}\le e\iff (n+p)\log\left(1+\frac1n\right)\le1 \iff p\le\frac1{\log\left(1+\frac1n\right)}-n.\tag3$$ Let us prove that the function $$f(x)=\frac1{\log\left(1+\frac1x\right)}-x$$ is increasing for $$x>0$$. Indeed $$f'(x)=\frac1{x(1+x)\left[\log\left(1+\frac1x\right)\right]^2}-1>0$$ in view of $$(2)$$.

Since $$f(n)$$ is increasing the largest possible $$p$$ corresponds to $$n=1$$, i.e. $$p=\frac1{\log2}-1\approx0.442695.$$

• Thank you very much, it’s a very clever solution! Mar 15, 2021 at 19:36
• You are welcome!
– user
Mar 15, 2021 at 19:54