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

Question

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?

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
@M.Noussa Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work — user, Mar 17 '18 at 22:58

score 1 · Answer 1 · answered Mar 13 '18 at 19:55

1

Note that

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

then use

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

answered Mar 13 '18 at 19:55

user

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

1 Answers1