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?
 A: Your algebra is cooky!
Note that
$$ \left( \frac{1}{2} \right)^2 \sum_{n=2}^\infty \left( \frac{1}{2} \right)^n = \sum_{n=2}^\infty \left( \frac{1}{2} \right)^{n+2} = \sum_{n=0}^\infty \left( \frac{1}{2} \right)^{n + 4}$$
A: You reindexed the sum incorrectly when you changed from $n$ to $k$ as dummy variable.
Just look at the first terms: when $n=2$, the term is $(\frac12)^2\cdot(\frac12)^2=\frac1{16}$; but when $k=0$, your first term is $(\frac12)^0=1$. Your first index for $k$ should be $4$, not $0$.
A: If the others' help was not sufficient, let me explain it to you like this:
You tried to calculate $S = \sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n.$ The first terms of this sum are $$S=\frac14 + \frac18 + \frac1{16}+\dots.$$You tried to do this by premultiplying by $1/4,$ probably because you wanted the first term to be $1$ to use the well-known expression $1 + x + x^2+\dots = 1/(1 - x)$. 
However: if you multiply both sides by $1/4$ the first term is instead $1/16,$ which is not $1$ and therefore not what you were trying for at all! In fact we should expect (since you meant to multiply by $4$ and instead multiplied by $1/4$) that you are off by a factor of $16$, and in fact you are: you got $8$ when you expected $1/2.$
We can use the corrected form of your reasoning to discover that $$\sum_{k=n}^\infty x^k = x^n \sum_{k=n}^\infty x^{k-n} = x^n \sum_{m=0}^\infty x^m = \frac{x^n}{1 - x},$$ where the "proper" substitution $m = k - n$ correctly maps our first index $k=n$ to the value $m = 0.$ 
In a little more detail: your error was to premultiply by $x^{-n}$ and substitute instead $m = k + n,$ both perfectly valid, but then you said that this resulting series starts from $m = 0$ when it does not: it starts from $m = 2n$ when we substitute $k=n$ into that relation for $m$ there.
