# Differentiation of summation

How do I show that \begin{align} \sum_{n=1}^\infty nz^{n-1} = \sum_{n=0}^{\infty} (n+1)z^n \end{align} (This is a result of differentiating $\sum_{n=1}^{\infty} z^n$ with respect to the variable $z$.)

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change the dummy variable $m\to n+1$. Write out the first several terms you can see that they are the same. – Arctic Char Mar 6 '14 at 9:43
Ahh they are both $1 +2z + 3z^2 + 4z^3 + \cdots$. – Cookie Mar 6 '14 at 9:52
You can deterministically know that the new summation should start at $0$ after replacing $n$ by $n+1$ in the summand. If it is not intuitive, you just have to make the same substitution down in the lower limit (and also in the upper limit.) – alex.jordan Mar 24 '14 at 7:12

Replace every instance of $n$ with $n+1$. When you write $$\sum_{n=1}^{M}$$ you really mean something like $$\sum_{n=1}^{n=M}$$ So you have something like $$\sum_{n=1}^{n=\infty}n\,z^{n-1}$$ and if we replace every instance of $n$ with $n+1$: $$\sum_{n+1=1}^{n+1=\infty}(n+1)\,z^{(n+1)-1}$$ which simplifies to $$\sum_{n=0}^{n=\infty}(n+1)\,z^{n}$$ (This is very similar to doing $u$-substitution on a definite integral with a shift ($x=u+c$) and also adjusting the limits of integration.)