<s>[INST] Does $c\cdot\sum\limits_{n=k}^{\infty}a_{n}= \sum\limits_{n=k}^{\infty}c\cdot a_{n}$ provided that the series converge? I am struggling to find what is wrong about this reasoning when calculating a series that does not start at $n=0$.
For instance, let $S = \sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n$.
Then $\left(\frac{1}{2}\right)^2S=\left(\frac{1}{2}\right)^2\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n$ (I)
And:
$$\left(\frac{1}{2}\right)^2S=\left(\frac{1}{2}\right)^2\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n=\sum\limits_{n=2}^{\infty}\left(\left(\frac{1}{2}\right)^n\cdot\left(\frac{1}{2}\right)^2\right)=\sum\limits_{k=0}^{\infty}\left(\frac{1}{2}\right)^k=2$$
But that means (from I): 
$$\left(\frac{1}{2}\right)^2S = 2 \Rightarrow \dfrac{\frac{1}{4}}{\frac{1}{4}}S=\dfrac{2}{\frac{1}{4}}\Rightarrow S = 8$$
Which is clearly wrong since:
$$\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n=\sum\limits_{n=0}^{\infty}\left(\frac{1}{2}\right)^n-\sum\limits_{n=0}^{1}\left(\frac{1}{2}\right)^n=\sum\limits_{n=0}^{\infty}\left(\frac{1}{2}\right)^n - 1 -\frac{1}{2}=\frac{1}{2}$$
I suspect that there is something behind this step $\left(\frac{1}{2}\right)^2\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n=\sum\limits_{n=2}^{\infty}\left(\left(\frac{1}{2}\right)^n\cdot\left(\frac{1}{2}\right)^2\right)$ that I am missing.
So my question is: 
Considering that $\sum\limits_{n=k}^{\infty}a_{n}$ is well defined (converges?), does:
$$c\cdot\sum\limits_{n=k}^{\infty}a_{n}= \sum\limits_{n=k}^{\infty}c\cdot a_{n}$$
And if it does, why does the reasoning presented above is wrong?
 [/INST] Your algebra is cooky!
Note that
$$ \left( \frac{1}{2} \right)^2 \sum_{n=2}^\infty \left( \frac{1}{2} \right)^n = \sum_{n=2}^\infty \left( \frac{1}{2} \right)^{n+2} = \sum_{n=0}^\infty \left( \frac{1}{2} \right)^{n + 4}$$
</s>