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 (;

It's simpler than that:
$$\frac{1}{n^2}-\frac{1}{(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}=\frac{2n+1}{n^2(n+1)^2}=\frac{2+1/n}{n(n+1)^2}\approx \frac{2}{n^3}$$
as for large $n$ one has $1/n\approx 0$ and $n+1\approx n$
