I am lost on this Taylor series. The hint says to let $x = 1/n$ and expand around $x = 0$, but I can't make any progress.
I am also confused why this hint is helpful. Can't I just expand around large $n$? Then I can assume $1/n^2, 1/n^3, ...$ and the other higher order derivatives to be about $0$, hence the Taylor series expansion would be $0$. Any advice?
On
Let us do it with $x=\frac 1n$ $$F(x)=\frac{(2-x) x^3}{(1-x)^2}$$ Now, use the binomial expansion for the denominator $$\frac{1}{(1-x)^2}=1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+7 x^6+\cdots$$ $$F(x)=(2-x) x^3 \left( 1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+7 x^6+\cdots\right)$$ $$F(x)=2 x^3+3 x^4+4 x^5+5 x^6+6 x^7+7 x^8+8 x^9+\cdots$$ $$F(x)=\sum_{k=2}^\infty k x^{k+1 }$$ $$F(n)=\sum_{k=2}^\infty \frac k {n^{k+1 }}$$
$\begin{array}\\ F(n) &= \dfrac1{(n-1)^2} - \dfrac1{n^2}\\ &= \dfrac1{n^2}\left(\dfrac1{(1-1/n)^2} -1\right)\\ &= \displaystyle\dfrac1{n^2}\left(\sum_{k=0}^{\infty}\dfrac{k+1}{n^{2k}} -1\right)\\ &= \displaystyle\dfrac1{n^2}\sum_{k=1}^{\infty}\dfrac{k+1}{n^{2k}} \\ &= \displaystyle\sum_{k=1}^{\infty}\dfrac{k+1}{n^{2k+2}} \\ &= \displaystyle\sum_{k=2}^{\infty}\dfrac{k}{n^{2k}} \\ \end{array} $