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,
$$a_n\leq S_n-\text{lower bound applied to the term $\left(1+\frac{1}{n-1}\right)^{n-1}$}\leq\frac{C}{n^2}$$
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.