Proving that $\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}$ when $n$ is very large This is an example from Mathematical Methods in the Physical Sciences, 3e, by Mary L. Boas.
My question is,
\begin{equation}
\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}
\end{equation}
can also be written as,
\begin{equation}\frac{1}{(-n)^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3} \end{equation}
so that $\Delta n = dn = -n-(n+1) = -2n-1$.
For $f(n) = 1/n^2$, $f'(n) = -2/n^3$, and
\begin{equation}
df = d(\frac{1}{n^2}) = f'(n)dn
\end{equation}
\begin{equation}
df = \frac{(2)(2n+1)}{n^3}
\end{equation}
Now, for very large $n$, $2n+1 \approx 2n$. Thus, \begin{equation}
df = \frac{4}{n^2}
\end{equation}
But, $4/n^2$ is not approximately equal to $2/n^3$ (required ans.) even if $n$ is very large.
So, please point out my mistake(s).
Thanks in advance (;

The claim can be shown with straightforward algebra as b00n heT points out. However, you could also use the mean value theorem, which formalizes your idea.
As for your approach, it's not clear to me why you wrote $\frac{1}{n^2}=\frac{1}{(-n)^2}.$ This is of course true, but this does not make $n=-n$, which is essentially what your subsequent step uses when you claim $\Delta n=-2n-1.$ Herein lies your error.
Let's tie the loose ends on your idea:
Defining $f(x)=1/x^2,$ the mean value theorem tells us
$$\underbrace{f(n+1)-f(n)}_{\Delta f}=f'(n+\epsilon_n)\underbrace{(n+1-n)}_{\Delta n}=f'(n+\epsilon_n),\quad \epsilon_n\in (0,1)\\
\implies \frac{1}{(n+1)^2}-\frac{1}{n^2}=-\frac{2}{(n+\epsilon_n)^3}\\
\implies \frac{1}{n^2}- \frac{1}{(n+1)^2}\sim \frac{2}{n^3},$$
where $g(n)\sim h(n)$ means $g(n)/h(n)\rightarrow 1$ as $n \uparrow \infty$.
Notice how your approach would get this result had you correctly used $\Delta n=1.$
