Q:

Shift Index on an Infinite Summation

I am learning about power series solution for differential equations and I have a question about shifting indexes and when you can bring a variable within the summation.
Specifically, how does:
$$2x \sum_{n=1}^{\infty} na_nx^{n-1} = 2\sum_{n=0}^{\infty} na_nx^{n}$$
My first thought was to bring x inside the summation on the LHS to get:
$$2 \sum_{n=1}^{\infty} na_nx^{n+1-1} = 2 \sum_{n=1}^{\infty} na_nx^{n}$$
then shifting indexes:
$$ 2 \sum_{n=1}^{\infty} na_nx^{n} = 2\sum_{n=0}^{\infty} (n+1)a_{n+1}x^{n+1}$$
Is this the same as the original equality? Because if you plug in $n=0$ to the original equality the first term is 0? If so, how can you justify removing the $+1$ term from the final equality?

A:

Your final result is equivalent, but to see why the summations are equivalent, simply multiply each term in the summation by $x$ to get that $2x\sum_{n=1}^\infty na_n x^{n-1}=2\sum_{n=1}^\infty na_n x^n$. When $n=0$ the first term is $0$ and it does nothing to the sum. Hence the result.

