Index shift in Taylor series of $\sin(x)$

Question

I saw the following index shift involving the Taylor Series for $\sin(x)$.
$$\sin(x)=\sum^{\infty}_{n=1}(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!}=\sum^\infty_{n=0}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$$ This second sum seems right to me, but the index shift seems funny. If I shifted the index down one – from $n=1$ to $n=0$, then shouldn't I shift all $n$'s in the equation up one? This would give me: $\sum^\infty_{n=0}(-1)^n\frac{x^{2x}}{2n!}$. Is the way I'm thinking about index shifts wrong?

Answer 1

You should consider that $2(n+1)-1=2n+1$.