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:

 A: Because after you have the expression $1+2r+3r^2+\ldots + nr^{n-1}$.
letting $r \to 1$, will reduce the expression to  $1+2+\ldots + n$ which is what you are interested.
A: (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 $$\lim_{r\to 1}1+2r+3r^2+\dots+nr^{n-1}=1+2+3+...+n$$
A: Here is the rationale behind Proof XI:
The formula
$$1+r+r^2+\ldots+r^n={1-r^{n+1}\over 1-r}\tag{1}$$
for the sum of a finite geometric series is given to us. It is valid for all $r\ne1$. We may differentiate $(1)$ on both sides as long as $r\ne1$:
$$1+2r+3r^2+\ldots+n r^{n-1}={\Psi(r)\over (1-r)^2}\qquad\forall r\ne1\ ,\tag{2}$$
where $\Psi(r)$ is a certain expression, given in your book. Now we observe that the LHS of $(2)$, evaluated at $r=1$, reduces to the sum $s_n:=1+2+3+\ldots+n$ which we are actually interested in. 
Unfortunately we cannot just evaluate the RHS of $(2)$ at $r=1$. Therefore we have to resort to limits: The LHS $f(r)$ is a continuous function of $r$ for all $r$. Therefore we may write
$$s_n=f(1)=\lim_{r\to1-}f(r)=\lim_{r\to1-}{\Psi(r)\over (1-r)^2}\tag{3}$$
because "during the limit process" we have $r\ne1$. The limit on the RHS of $(3)$ can indeed be calculated, because de l'Hôpital comes twice to the rescue, and we obtain the expected result.
