# How to calculate the sum of $\sum_{n=0}^\infty (n-4)\cdot\left(\frac{1}{2}\right)^n$ [duplicate]

This sum: $$\sum_{n=0}^\infty n\cdot\left(\frac{1}{2}\right)^{n-1}$$ can be calculated using this $\sum_{n=0}^\infty a^n=\frac{1}{1-a}$ ; $a<1$ cause we can calculate the sum $\sum_{n=0}^\infty \left(\frac{1}{2}\right)^{n}$ and then calculate it's derivative.

Can we calculate this sum: $$\sum_{n=0}^\infty (n-4)\cdot \left(\frac{1}{2}\right)^n$$ using a similar method?

If not, how can I calculate it?

## marked as duplicate by Jyrki LahtonenMar 13 '18 at 21:23

• Hint: \begin{eqnarray*} \sum_{n=0}^{\infty} n x^n =\frac{x}{(1-x)^2}. \end{eqnarray*} – Donald Splutterwit Mar 13 '18 at 19:39
• thank you so much! i got it. – M.Noussa Mar 13 '18 at 19:40
• @M.Noussa Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… – gimusi Mar 17 '18 at 22:58

$$\sum_{n=0}^\infty (n-4)\cdot \left(\frac{1}{2}\right)^n=\sum_{n=0}^\infty n\cdot \left(\frac{1}{2}\right)^n-4\sum_{n=0}^\infty \left(\frac{1}{2}\right)^n$$
$$\sum_{n=0}^{\infty} n x^n =\frac{x}{(1-x)^2}$$