Find $Q''\left(\frac{3}{2}\right)$ where $Q(x) = 5-3(x-2)+\sum_{n=2}^{\infty}\frac{(x-2)^n}{n(n-1)}$

Question

Find $Q''\left(\frac{3}{2}\right)$ where $Q(x) = 5-3(x-2)+\sum_{n=2}^{\infty}\frac{(x-2)^n}{n(n-1)}$. I think I am supposed to use Taylor series, but in what way?

Answer 1

Hint $${d^2\over dx^2}{(x-2)^n\over n(n-1)}=(x-2)^{n-2}\quad,\quad n>1$$

Answer 2

$$Q(x) = 5-3(x-2)+\sum_{n=2}^{\infty}\frac{(x-2)^n}{n(n-1)}$$ Differentiate $$Q'(x) = -3+\sum_{n=2}^{\infty}\frac{(x-2)^{n-1}}{(n-1)}$$ Differentiate again: $$Q''(x) = \sum_{n=2}^{\infty}{(x-2)^{n-2}}$$ Change the index of summation: $$Q''(x) = \sum_{n=0}^{\infty}{(x-2)^{n}}$$ $$Q''\left(\frac 3 2 \right) = \sum_{n=0}^{\infty}{\left(-\dfrac 12 \right)^{n}}$$ Apply the Taylor formula for $|x|<1$ : $$\sum_{n=0}^{\infty}x^{n}=\dfrac 1 {1-x}$$

Answer 3

Note that taking the second derivative of $Q(x)$ causes constant and linear terms to disappear thus: \begin{align*} Q''(x) &= \sum_{n=2}^{\infty} \dfrac{n(n-1)\cdot (x-2)^{n-2}}{n(n-1)} \\ &= \sum_{n=2}^{\infty} (x-2)^{n-2}. \end{align*}Therefore $Q''$ is given by a geometric series, so $$Q''\left(\dfrac{3}{2}\right)=\sum_{n=2}^{\infty} \left(-\dfrac{1}{2}\right)^{n-2} = \dfrac{1}{1-\left(-\dfrac{1}{2}\right)} = \dfrac{1}{~3/2~} = \boxed{\dfrac{2}{3}}.$$