Another way to find the sum $S=1-\dfrac{3}{2}+\dfrac{5}{4}-\dfrac{7}{8}+\cdots+(-1)^{n-1}\cdot\dfrac{2n-1}{2^{n-1}}$

Question

I want to know the sum $$S=1-\dfrac{3}{2}+\dfrac{5}{4}-\dfrac{7}{8}+\cdots+(-1)^{n-1}\cdot\dfrac{2n-1}{2^{n-1}}.$$ This is my way. First, we find the sum $$1 + 3 x + 5x^2 + \cdots + (2n-1) \cdot x^{n-1}=\sum _{k=1}^n (2 k-1) x^{k-1}.$$ We have $$\sum _{k=1}^n (2 k-1) x^{k-1} = 2\sum _{k=1}^n k\cdot x^{k-1}-\sum _{k=1}^n x^{k-1}. $$ Note that $$\sum _{k=1}^n k\cdot x^{k-1} = \left (\sum _{k=1}^n x^k\right)' = \left (\dfrac{x\left(x^n-1\right)}{x-1}\right)'=\dfrac{n x^{n+1}-(n+1)x^n+1}{(x-1)^2}.$$ Another way $$\sum _{k=1}^n x^{k-1} = \dfrac{x^n-1}{x-1}.$$ From the above results, we have $$1 + 3 x + 5x^2 + \cdots + (2n-1) \cdot x^{n-1}= \dfrac{(2 n-1) x^{n+1}-(2 n+1)x^n+x+1}{(x-1)^2}.$$ With $x=-\dfrac{1}{2}$,we get $$S=\dfrac{2^n + (-1)^{n+1} \cdot(6n+1)}{9 \cdot 2^{n-1}}.$$ How to find the result with another way?

Answer 1

$$S+\frac S2=\sum_1^n\frac{(-1)^{k-1}(2k-1)}{2^{k-1}}+\sum_1^n\frac{(-1)^{k-1}(2k-1)}{2^{\color{red}k}}\\ =\sum_1^n\frac{(-1)^{k-1}(2k-1)}{2^{k-1}}+\sum_\color{red}2^{\color{red}{n+1}}\frac{(-1)^{k-2}(2k-3)}{2^{\color{red}{k-1}}}\\ =1+\sum_2^n\frac{(-1)^{k-2}(\color{red}{2k-1-2k+3})}{2^{k-1}}+\frac{(-1)^{n-1}(2n-1)}{2^n}.$$ The rest is well-known.

Answer 2

I would have done it in a similar but different way. Consider the polynomial$$1-3x^2+5x^4-7x^6+\cdots+(-1)^{n-1}(2n-1)x^{2n-2}.$$If its sum is denoted by $f(x)$ and if$$F(x)=x-x^3+x^5-x^7+\cdots+(-1)^{n-1}x^{2n-1},$$then $F'=f$. On the other hand$$F(x)=x\left(1-x^2+x^4-x^6+\cdots+(-1)^{n-1}x^{2n-2}\right)=\frac{x-(-1)^nx^{2n}}{1+x^2}.$$Therefore,$$f(x)=\frac{-(-1)^n (2 n+1) x^{2 n}+(-1)^n (1-2 n) x^{2n+2}-x^2+1}{(1+x^2)^2}$$and, in particular, your sum is$$f\left(\sqrt{\frac12}\right)=\frac{1}{9} 2^{1-n}\bigl(2^n-6 (-1)^n n-(-1)^n\bigr).$$

Answer 3

The general term of your series is on the form $(-1)^{n-1}\frac{2n-1}{2^{n-1}} = -4nx^n + 2x^n$ where $x = -\frac{1}{2}$. When summed the latter term gives rise to a geometrical series and for the former note that $$(1-x)\left[x + 2x^2 + 3x^3 + 4x^4 + \ldots + nx^n\right] = x+x^2+x^3+\ldots+x^{n} - nx^{n+1}$$ which is just another geometrical series. This gives you

$$\sum_{k=1}^n-4kx^k + 2x^k = -4\frac{\left(g_n(x) - nx^{n+1}\right)}{1-x} + 2g_n(x)$$

where $g_n(x) = x+x^2+\ldots+x^n =\frac{x-x^{n+1}}{1-x}$. Taking $x=-1/2$ gives the same result as you have found already.