2020-10-19

a) Find the Maclaurin series for the function
$f\left(x\right)=\frac{1}{1}+x$
b) Use differentiation of power series and the result of part a) to find the Maclaurin series for the function
$g\left(x\right)=\frac{1}{\left(x+1{\right)}^{2}}$
c) Use differentiation of power series and the result of part b) to find the Maclaurin series for the function
$h\left(x\right)=\frac{1}{\left(x+1{\right)}^{3}}$
d) Find the sum of the series
$\sum _{n=3}^{\mathrm{\infty }}\frac{n\left(n-1\right)}{2n}$
This is a Taylor series problem, I understand parts a - c but I do not understand how to do part d where the answer is $\frac{7}{2}$

Nathalie Redfern

It took me a while to get this, but finally I got it. Had to try a lot of tricks to get this series.
We have the series for $\frac{1}{\left(x+1{\right)}^{3}}$ as shown below:
$f\left(x\right)=\frac{1}{\left(1+x{\right)}^{3}}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2}\left(-1{\right)}^{n}\left(n+1\right)\left(n+2\right){x}^{n}$
We can similarly write a series for $\frac{1}{\left(1-x{\right)}^{3}}$ as shown below:
$f\left(x\right)=\frac{1}{\left(1-x{\right)}^{3}}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2}\left(n+1\right)\left(n+2\right){x}^{n}$
Now, since the index of the summation in our question starts from n=3, we could shift the index of this series to n=3. I tried that, but it didnt

Jeffrey Jordon