Summing a Sequence

Question

I recently came across a problem that proved particularly challenging. Unfortunately, I have not been able to complete the problem despite multiple different methods tried. It is as below:

Show that the sum of the series

$$1+3\left(\frac{1}{2}\right)+5\left(\frac{1}{2}\right)^2+7\left(\frac{1}{2}\right)^3+...+\left(2n-1\right)\left(\frac{1}{2}\right)^{n-1}$$

is $6-R_n$ where $$R_n=\frac{2n+3}{2^{n-1}}$$

I have tried the method mentioned.

$$\sum_{i=0}^{\infty}{x^i}=\frac{1}{1-x}$$ $$\sum_{i=0}^{\infty}ix^{i-1}=\frac{1}{(1-x)^2}$$ $$\sum_{i=0}^{\infty}ix^i=\frac{x}{(1-x)^2}$$ $$\sum_{i=0}^{\infty}(2i-1)x^{i-1}=\frac{x+1}{(1-x)^2}$$ $$\sum_{i=0}^{\infty}(2i-1)\left(\frac{1}{2}\right)^{i-1}=\frac{\frac{1}{2}+1}{\left(1-\frac{1}{2}\right)^2}$$ $$=6$$

However, the sum is finite, so this method cannot be used. I have attempted a similar approach with the finite geometric sequence sum, but got an incorrect result.

I am grateful for any help. Thanks.

Answer 1

You have to calculate finite sums as below: $$ S=\sum_{k=1}^n(2k-1)x^{k-1}. $$ Hence, $$ S=2\sum_{k=1}^nkx^{k-1}-\sum_{k=1}^nx^k $$ But $$ \sum_{k=1}^nx^k=\frac{1-x^{k+1}}{1-x}-1=\frac{x(1-x^k)}{1-x} $$ and $$ \sum_{k=1}^nkx^{k-1}=\frac{d}{dx}\left(\sum_{k=1}^nx^k\right)=\frac{d}{dx}\left(\frac{x(1-x^k)}{1-x}\right). $$ Consequently, $$ S=2\frac{d}{dx}\left(\frac{1-x^{k+1}}{1-x}\right)-\left(\frac{x(1-x^k)}{1-x}\right). $$ By making $x=1/2$ the desired result is obtained.

Answer 2

Hint: the sum of the geometric series $$\sum_{i=0}^\infty a^i=\frac {1}{1-a}$$ If you differentiate with respect to $a$ you get $$\sum_{i=0}^\infty ia^i=\frac 1{(1-a)^2}$$ You need to deal with the fact that your series is finite. You can set $a=\frac 12$ in the end.

Answer 3

\begin{align}\sum_{i=0}^{n-1}(2i+1) \left( \frac12\right)^i&= \sum_{i=0}^{n-1}i\left( \frac12\right)^{i-1}+ \sum_{i=0}^{n-1}\left( \frac12\right)^i\\ &=\sum_{i=0}^{n-1} \left.\frac{d}{dx}x^i\right|_{x=0.5}+\frac{1-\frac1{2^n}}{\frac12} \\ &=\frac{d}{dx}\left.\left(\frac{1-x^{n}}{1-x}\right)\right|_{x=0.5}+2(1-\frac1{2^n})\\ &=\left.\frac{(1-x)(-nx^{n-1})+(1-x^{n})}{(1-x)^2}\right|_{x=0.5}+2(1-\frac1{2^n})\\ &=\frac{\frac12(-\frac{n}{2^{n-1}})+(1-\frac1{2^{n}})}{\frac14}+2-\frac1{2^{n-1}}\\ &=2(-\frac{n}{2^{n-1}})+4-\frac2{2^{n-1}}+2-\frac1{2^{n-1}}\\ &= 6-\frac{2n+3}{2^{n-1}}\end{align}