A simple but nonsense infinite sums rule Assume $a>0$ and $0<r<1$ are given. The geometric series is known to be $\sum_0^{\infty} ar^n = \frac{a}{1-r}$.  Now differentiate w.r.t $r$ to obtain $\sum_0^{\infty}anr^n = \frac{ar}{(1-r)^2}$. So far we're cool.
Now take the two series $$S_1 = \sum_n \frac{1}{2^n}$$ and $$S_2 = \sum_n \frac{n}{2^n}.$$ One would trivially expect $S_2>S_1$, because obviously $S_2$ has an additional $n$ multiplied by each term. To my surprise, this is not the case! Invoking the result of the previous paragraph, you get
$$S_1 = \frac{1}{1-1/2} = 2$$
and
$$ S_2 = \frac{1/2}{1/4} = 2$$
so that $S_1 = S_2$.
Does this make any sense to you? Because it sure makes no sense to me. Or am I losing it?
 A: Yes, you are right $S_1 = S_2 = 2$ and it might be somewhat astonishing the first time. But first let's have a look at your derivation which needs to be somewhat revised. We can take $a=1$ and consider
\begin{align*}
S_1(r)=\sum_{n=0}^\infty r^n=\frac{1}{1-r}\tag{1}
\end{align*}
Derivation with respect to $r$ gives
\begin{align*}
\frac{d}{dr}S_1(r)&=\frac{d}{dr}\frac {1}{1-r}=\frac{1}{(1-r)^2}\tag{2}
\end{align*}
Note we have numerator $1$ and not $r$. On the other hand we also obtain
\begin{align*}
\frac{d}{dr}S_1(r)&=\frac{d}{dr}\sum_{n=0}^\infty r^n=\sum_{n=0}^\infty nr^{n-1}=\sum_{n=1}^\infty nr^{n-1}\\
&=\sum_{n=0}^\infty (n+1)r^n\\
&=\sum_{n=0}^\infty nr^n+\sum_{n=0}^\infty r^n\\
&=\sum_{n=0}^\infty nr^n+\frac{1}{1-r}\\
\sum_{n=0}^\infty nr^n& = \frac{d}{dr}S_1(r) - \frac{1}{1-r}=\frac{1}{(1-r)^2}-\frac{1}{1-r}\tag{3}
\end{align*}
We obtain from (1) - (3) by substituting $r=\frac{1}{2}$:
\begin{align*}
\color{blue}{S_1\left(\frac{1}{2}\right)}&=\frac{1}{1-\frac{1}{2}}\,\color{blue}{=2}\\
\color{blue}{S_2}&=\sum_{n=0}^\infty \frac{n}{2^n}
=\frac{1}{\left(1-\frac{1}{2}\right)^2}-\frac{1}{1-\frac{1}{2}}=4-2\,\,\color{blue}{=2}
\\
\end{align*}
We observe the term with $n=0$ in $S_2$ is zero, but the numerator $n$ with index $n>0$ is a compensation for it.
