Index of summation shift I'm learning about power series in Differential Equations.
Right now I'm learning about shifting summations and something that is bothering me is the following:
Take the equation
$$F(x) = (x-3)y' + 2y = 0$$
then substitute 
$$(x-3)\sum\limits_{n=1}^\infty c_nx^{n-1} +2\sum\limits_{n=0}^\infty c_nx^n$$
afterwards it gets simplified to 
$$ \sum\limits_{n=1}^\infty nc_nx^n -3\sum\limits_{n=1}^\infty nc_nx^{n-1} +2\sum\limits_{n=0}^\infty c_nx^n$$
then obtain
$$ \sum\limits_{n=0}^\infty nc_nx^n -3\sum\limits_{n=0}^\infty {(n+1)}c_{n+1}x^{n} +2\sum\limits_{n=0}^\infty c_nx^n$$
What is confusing me is why can the first summations $n$ be replaced to 0 without any effect on the sum? but in the second summation you have to shift everything by +1.
Anyone mind clearing it up for me? Thanks
 A: Because 
$$ \sum\limits_{n=0}^\infty nc_nx^n = 0\cdot c_0 \cdot x^0+\sum\limits_{n=1}^\infty nc_nx^n=\sum\limits_{n=1}^\infty nc_nx^n$$
Edit:
In the second sum it's actually also 
$$\sum\limits_{n=0}^\infty nc_{n}x^{n-1}=0\cdot c_0\cdot x^{-1}+\sum\limits_{n=1}^\infty nc_{n}x^{n-1}=\sum\limits_{n=1}^\infty nc_{n}x^{n-1}$$
But you also want to make the $x^{n-1}$ to have the same exponent as in the other sums, $x^n$. So you have to set $m=n-1$, so $n=m+1$ and therefore 
$$\sum\limits_{n=1}^\infty nc_{n}x^{n-1}=\sum\limits_{m=0}^\infty (m+1)c_{m+1}x^{m}$$
A: it would be possible, but the goal is to have $x$ to the power of $n$ everywhere, so as to compare the coefficients. note that you're missing that factor of $n$ in your second equation. originally the summation started from $n=0$:
$$
y(x) = \sum_{n=0}^{\infty}{c_n x^n}
$$
as to the derivative:
$$
y'(x) = \sum_{n=0}^{\infty}{n c_n x^{n-1}}
$$
the initial term (which is $0$ anyway) was dropped in order to be able to shift the exponent
A: I think the explanation this question is asking for is
$$\sum_{n=1}^\infty n c_n x^n = \big(\sum_{n=1}^\infty n c_n x^n\big) + 0 c_0 x^0 = \sum_{n = 0}^\infty n c_n x^n.$$
In other words, simply add the zero term, it is the constant term which when you took the derivative "vanished" and is the reason for the shift in index in the first place. Think about taking the derivative term by term of your supposed solution, then multiply it by x without shifting the index of the original sum.
