When taking derivatives of power series, why do we shift the index up? For example, if the series starts at n=0, and we take the derivative, the index usually then starts at n=1. This increases as we continue taking derivatives, but why do we need to do this? 
I get that if we don't the first term will be 0, but why not include them? Is this simply for convenience of not having to deal with the first few 0 terms? 
 A: Let $$A(x) = \sum_{n=0} ^ \infty a_nx^n = a_0 + a_1 x + a_2 x^2 + \cdots$$ be a serie (centered in 0 for this example, if centered in $z$ it would be $\sum_{n=0} ^ \infty a_n(x-z)^n$). Then, if we formally derivate on $x$ (like in a polynomial) we get $$A'(x) = a_1 + 2 a_2 x + 3a_3 x^2 + \cdots = \sum_{n=0} ^ \infty (n+1)a_{n+1}x^n.$$
Then it gets a bit messy for the reader, so we can do $k = n+1$ and: 


*

*When $n=0$ we have $k=1$.

*Also we need to know the value of $n$ with respect to $k$, that is $n=k-1$.

*Obviously the upper limit $\infty$ is not affected by this change of variables.


Now we can finally rewrite the serie as: 
$$A'(x) = \sum_{k=1} ^ \infty ka_{k}x^{k-1}.$$
This way we are clearly saying that the first term will be $a_1$. If we would write $A'(x) = \sum_{k=0} ^ \infty ka_{k}x^{k-1}$ the reader needs to realize that the first term is multiplied by 0 so its null...
For the $i$-th derivative it can be shown (by induction) that 
$$A^{(i)}(x) = \sum_{k=i} ^ \infty (k·(k-1)\cdots(k-i+1))a_{k}x^{k-i},$$
with $k=n+i$. Here, if we would write $A^{(i)}(x) = \sum_{k=0} ^ \infty (k·(k-1)\cdots(k-i+1))a_{k}x^{k-i}$, when $k=0$ the term is 0, when $k=1$ the term is 0 cause $k-1$ is 0 and so on... Hard and messy to know when the series truly starts, right?
Hope that clarifies it a bit.
A: It is perfectly legit (and common, I don't know what has influenced your perception) to reindex as you see fit:
$$\frac{d}{dx}\sum_{k=0}^\infty a_k x^k = \sum_{k=1}^\infty k a_k x^{k-1}= 
\sum_{k=0}^\infty(k+1) a_{k+1} x^k
$$ 
