Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $z$.)

share|cite|improve this question
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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.