I wish to sum the series $\sum_{n=1}^{\infty}\dfrac{2n-1}{2^n}$.
I notice that by writing $\displaystyle\sum_{n=1}^{\infty}(2n-1)\dfrac{1}{2^n}$, $\dfrac{1}{2^n}$ is a geometric series and can be summed easily. But the next step I think I did wrong is
$$2n-1\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{2^n}$$
I don't think I can pull out $2n-1$ like a constant because that will affect the sum. What should I have done. The series converges to $3$ according Wolfram Alpha.
Best Answer
Here is one way: $$ \sum_{n=1}^\infty \frac{2n-1}{2^n}=\sum_{n=1}^\infty \frac{2n}{2^n}-\sum_{n=1}^\infty\frac{1}{2^n}. $$ The second sum is geometric and converges to $1$. We just need to deal with the first. To do this, use the relation $$ \sum_{n=0}^\infty z^n=\frac{1}{1-z} $$ valid for $\lvert z\rvert <1$ and differentiate to get $$ \sum_{n=1}^\infty nz^{n-1}=\frac{1}{(1-z)^2}. $$ Next, evaluate at $z=\frac{1}{2}$ to get $$ \sum_{n=1}^\infty \frac{2n}{2^n}=\frac{1}{(1/2)^2}=4. $$ So, the result is $3$.