# The sum of the series by derivatives

We have a series $$\sum_{k=3}^\infty k \frac{\bigl(\frac{1+\sqrt5}{2}\big)^{k-1}}{2^{k-1}}$$ and I want to compute it like that. Let's say that $$a = \bigl(\frac{1+\sqrt5}{2}\bigr)^{k-1}$$. Then we have $$\sum_{k=3}^\infty k \frac{a^{k-1}}{2^{k-1}} = \sum_{k=3}^\infty k (\frac{a}{2})^{k-1}$$ I tried to rewrite it as a derivative: \begin{align} \sum_{k=3}^\infty k \Bigl(\frac{a}{2}\Bigr)^{k-1} &= \sum_{k=3}^\infty \frac{d(\frac{a}{2})^{k}}{da} \\ &= \frac{d\frac{(\frac{a}{2})^{3}}{(1 - \frac{a}{2})}}{da} \\ &= \frac{3 * (\frac{a}{2})^2 * (1 - \frac{a}{2}) + (\frac{a}{2})^3 * \frac{1}{2}}{(1 - \frac{a}{2})^{2}} \end{align} But the answer is wrong. What did I do incorrectly?

Your idea of differentiating a geometric series was good (and standard, but I couldn't find exact duplicates on this site) but you forgot twice the factor $$\frac12$$ when applying the chain rule: \begin{align}\color{red}{\frac12}\sum_{k=3}^\infty k\left(\frac a2\right)^{k-1}& = \sum_{k=3}^\infty \frac{d\left(\frac a2\right)^k}{da} \\&= \frac{d\frac{(\frac a2)^3}{1 - \frac a2}}{da} \\&= \frac{3 (\frac a2)^2 (1 - \frac a2) \color{red}{\frac12}+ (\frac a2)^3 \frac12}{(1 - \frac a2)^2}\end{align} You would have avoided these mistakes if you let $$x=\frac a2=\frac{1+\sqrt5}4:$$ \begin{align}\sum_{k=3}^\infty kx^{k-1}& = \sum_{k=3}^\infty \frac{dx^k}{dx} \\&= \frac{d\frac{x^3}{1 -x}}{dx} \\&= \frac{3x^2 (1 -x)+x^3}{(1 -x)^2}\\&= \frac{3x^2-2x^3}{(1 -x)^2}.\end{align}
Let $$a=\frac{1+\sqrt5}{2×2}$$ So your summation becomes $$\sum_{k=3}^{\infty}ka^{k-1}$$ Now we will unwrap this summation. Let $$S=3a^2+4a^3+5a^4+\cdots$$ $$\implies S\cdot a=3a^3+4a^4+\cdots$$ $$S-Sa=3a^2+a^3+a^4+\cdots$$ We didn't touch the first term of the $$S$$ and subtracted terms of same literal coefficient. Now $$S(1-a)=2a^2+(a^2+a^3+a^4+\cdots)$$ $$S(1-a)=2a^2+\frac{a^2}{1-a}$$ $$\implies S=\frac{3a^2-2a^3}{(1-a)^2}$$ Now undo the substitution to get the final answer.