Series expansion from the red book on special functions by Richard Ask

[MathematicalPhysicist]

I want to check my calculations via mathematica.

In the book I am reading there's this expansion:
$$\frac{(1+\frac{1}{j})^x}{1+x/j}=1+\frac{x(x-1)}{2j^2}+\mathcal{O}(1/j^3)$$

though I get instead of the term $\frac{x(x-1)}{2j^2}$ in the rhs the term: $-\frac{x(x+1)}{2j^2}$.

So I want to check by mathematica if my calculations are correct, how do you suggest me to implement it in mathematica?

Thanks!

 

[stevendaryl]

I don't have experience with Mathematica, but I suggest you try the example x=2 to see if your expansion is right.

 

[DrClaude]

Define $k \equiv 1/j$ then use the command Series to get the series expansion about $k=0$.

Spoiler

The book is correct.

 

[MathematicalPhysicist]

Define $k \equiv 1/j$ then use the command Series to get the series expansion about $k=0$.

Spoiler

The book is correct.

Yeah I know.
I thought to myself to expand $(1+1/j)^x = 1+x/j+x(x-1)/(2j^2)+\mathcal{O}(1/j^3)$ and $(1+x/j)^{-1}=1-x/j+\mathcal{O}(1/j^2)$ and then multiply both factors; but it seems not to be the correct approach.

 

[pasmith]

Yeah I know.
I thought to myself to expand $(1+1/j)^x = 1+x/j+x(x-1)/(2j^2)+\mathcal{O}(1/j^3)$ and $(1+x/j)^{-1}=1-x/j+\mathcal{O}(1/j^2)$ and then multiply both factors; but it seems not to be the correct approach.

You need to expand both to order $j^{-2}$: $$\begin{align*} \left(1 + \frac 1j\right)^x &= 1 + \frac xj + \frac{x(x-1)}{2j^2} + O(j^{-3}) \\ \left(1 + \frac xj \right)^{-1} &= 1 - \frac xj + \frac{x^2}{j^2} + O(j^{-3}) \end{align*}.$$ Now multiply these together, ignoring any term of order $j^{-3}$ or higher: $$\begin{split} \left(1 - \frac xj + \frac{x^2}{j^2} + O(j^{-3})\right) + \frac xj \left( 1 - \frac xj + O(j^{-2}) \right) + \frac{x(x-1)}{2j^2} \left(1 + O(j^{-1})\right) \\= 1 + \frac{x(x-1)}{2j^2} + O(j^{-3}).\end{split}$$ Omitting $+x^2/j^2$ from the first bracket will give the incorrect $$1 - \frac{x(x+1)}{2j^2} + O(j^{-3})$$ from your original post.