The maximum value of $p>0$ such that $(1+\frac{1}{n})^{n+p}Find the maximum value of $p>0$ such that $(1+\frac{1}{n})^{n+p}<e$ 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?
 A: 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.
$$
p=\frac1{\log2}-1\approx0.442695.
$$
