Bounding the difference between consecutive terms in the sequence $a_n=(1+1/n)^n$ I'm pretty sure the following estimate holds for $C>1$, but I can't prove it:
$$a_n:=\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n-1}\right)^{n-1}\leq\frac{C}{n^2}$$
For context, I was looking for some such bound in order to prove that $na_n\to0$.
I tried using the fact that
$$\left(1+\frac{1}{n}\right)^n\leq\color{red}{1}+\color{red}{1}+\color{red}{\frac{1}{2}}+\cdots+\color{red}{\frac{1}{n!}}:=S_n$$
because, by some algebra with the binomial theorem,
$$\left(1+\frac{1}{n}\right)^n=\color{red}{1}+\color{red}{1}+\color{red}{\frac{1}{2}}\color{blue}{\left(1-\frac{1}{n}\right)}+\cdots+\color{red}{\frac{1}{n!}}\color{blue}{\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{n-1}{n}\right)}$$
where the blue factors are all less than one.
But I also need a good lower bound on the terms $(1+n^{-1})^{n}$, and I can't see how to pick the right lower bound to get, say,
$$a_n\leq S_n-\text{lower bound applied to the term $\left(1+\frac{1}{n-1}\right)^{n-1}$}\leq\frac{C}{n^2}$$
 A: If you use Taylor series
$$b_p=\left(1+\frac{1}{p}\right)^p\implies \log(b_p)=p\log\left(1+\frac{1}{p}\right)$$
$$\log(b_p)=p\Bigg[\frac{1}{p}-\frac{1}{2 p^2}+\frac{1}{3 p^3}-\frac{1}{4
   p^4}+\frac{1}{5 p^3}+O\left(\frac{1}{p^6}\right) \Bigg]$$
$$\log(b_p)=1-\frac{1}{2 p}+\frac{1}{3 p^2}-\frac{1}{4p^3}+\frac{1}{5 p^4}+O\left(\frac{1}{p^5}\right) $$
$$b_p=e^{\log(b_p)}=e-\frac{e}{2 p}+\frac{11 e}{24 p^2}-\frac{7 e}{16
   p^3}+\frac{2447 e}{5760 p^4}+O\left(\frac{1}{p^5}\right) $$
Apply is twice and continue with Taylor series to make
$$a_n=\frac{e}{2 n^2}-\frac{5 e}{12 n^3}+\frac{7 e}{16
   n^4}+O\left(\frac{1}{n^5}\right) $$
So, to make the inequality true for all $n$, we need $C >\frac{1}{2}e$
A: Suppose that $p\geq 1$. Then
$$
\left( {1 + \frac{1}{p}} \right)^p  = \exp \left( {p\log \left( {1 + \frac{1}{p}} \right)} \right) \ge \exp \left( {1 - \frac{1}{{2p}}} \right) \ge e\left( {1 - \frac{1}{{2p}}} \right)
$$
and
\begin{align*}
&\left( {1 + \frac{1}{p}} \right)^p  \le \exp \left( {p\log \left( {1 + \frac{1}{p}} \right)} \right)\le \exp \left( {1 - \left( {\frac{1}{{2p}} - \frac{1}{{3p^2 }}} \right)} \right)  \\ & \le e\left( {1 - \left( {\frac{1}{{2p}} - \frac{1}{{3p^2 }}} \right) + \frac{1}{2}\left( {\frac{1}{{2p}} - \frac{1}{{3p^2 }}} \right)^2 } \right) \le e\left( {1 - \frac{1}{{2p}} + \frac{{11}}{{24p^2 }}} \right).
\end{align*}
These inequalities follow from the fact that for all $x>0$,
$$
x - \frac{{x^2 }}{2} < \log (1 + x) < x - \frac{{x^2 }}{2} + \frac{{x^3 }}{3},\quad e^{ - x}  < 1 - x + \frac{{x^2 }}{2}.
$$
In the last step, I re-arranged the terms and observed that when multiplied by $p^2$, the sum of the terms beyond $-\frac{1}{2p}$ tend monotonically from below to $\frac{11}{24}$. Thus, for all $n\geq 2$,
\begin{align*}
\left( {1 + \frac{1}{n}} \right)^n  - \left( {1 + \frac{1}{{n - 1}}} \right)^{n - 1} & \le e - \frac{e}{{2n}} + \frac{{11}}{{24n^2 }} - e + \frac{e}{{2(n - 1)}}
\\ &
 = \frac{e}{2}\frac{1}{{n(n - 1)}} + \frac{{11}}{{24n^2 }} \le \frac{{24e + 11}}{{24}}\frac{1}{{n^2 }} < \frac{4}{{n^2 }} .
\end{align*}
Addendum: By the mean value theorem
\begin{align*}
&\left( {1 + \frac{1}{n}} \right)^n  - \left( {1 + \frac{1}{{n - 1}}} \right)^{n - 1}  = \left( {1 + \frac{1}{\xi }} \right)^\xi  \left( {\log \left( {1 + \frac{1}{\xi }} \right) - \frac{1}{{\xi  + 1}}} \right)
\\ &
 \le e\left( {\log \left( {1 + \frac{1}{\xi }} \right) - \frac{1}{{\xi  + 1}}} \right) \le \frac{e}{{2\xi ^2 }} \le \frac{e}{{2(n - 1)^2 }},
\end{align*}
where $n-1<\xi<n$ is a suitable number.
