In showing integer sum $(1+2+3+...+n)$ by l'Hopital rule why they take lim as $r$ approaches to $1$ in one of steps? Why $1$? So why lim as $r \to 1$ (why $1$?) Here's the method:


(Arithmetic - geometric progression)
If we differentiate both sides with respect to $r$, we have
\begin{equation}
1+2r+3r^2+\dots+nr^{n-1}=\frac{nr^{n+2}-(n+1)r^{n+1}+r}{(1-r)^2}
\end{equation}
Letting $r \to 1$ and applying L'Hospital's rule on the fraction, we end up with
\begin{equation}
\frac{n(n+1)}2=1+2+3+\dots+n
\end{equation}
$$\lim_{r\to 1}\frac{nr^{n+2}-(n+1)r^{n+1}+r}{(1-r)^2}=\lim_{r\to 1}1+2r+3r^2+\dots+nr^{n-1}\\
\lim_{r\to 1}\frac{nr^{n+2}-(n+1)r^{n+1}+r}{(1-r)^2}=\frac{0}{0} =\\
\lim_{r\to 1}\frac{n(n+2)r^{n+1}-(n+1)^2r^{n}+1}{(2(1-r)(-1))}=\frac{0}{0}\\\
\lim_{r\to 1}\frac{n(n+2)(n+1)r^{n}-(n+1)^2nr^{n-1}+0}{(2(-1)(-1))}=\\
\frac{n(n+2)(n+1)-n(n+1)^2}{+2}=\\
\frac{n((n^2+3n+2-(n+1)^2)}{+2}=\\
\frac{n(n+1)}{2}$$R..H.S. is $$\lim_{r\to 1}1+2r+3r^2+\dots+nr^{n-1}=1+2+3+...+n$$
