Can somebody tell me how the Taylor expansion of$$\ln\left(\frac{x}{n}+1\right)=\frac{x}{n}-\frac{x^2}{2n^2}+\frac{x^3}{3n^3}-\cdots?$$

According to the definition,$$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots,$$but I can't seem to make the connection.

Question

Can somebody tell me how the Taylor expansion of$$\ln\left(\frac{x}{n}+1\right)=\frac{x}{n}-\frac{x^2}{2n^2}+\frac{x^3}{3n^3}-\cdots?$$

According to the definition,$$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots,$$but I can't seem to make the connection.

anonymous · Answer

i can come back to this but if you want an immediate guess i would say take the derivative, 
probably get an easy expansion, and then integrate term by term
that is just a guess though

anonymous · Answer

ok i am confused 
why does your function have an "n" in it??

anonymous · Answer

damn it is not working. i thought you could integrate this thing term by term and get what you want, but maybe i made a calculation error somewhere

anonymous · Answer

we could try the expansion as you wrote it
we are expanding about 0 so 
$$f(0)=\ln(1)=0$$ first term is 0

$$f'(0)=\frac{n}{n+0}=1$$ so second term is x

anonymous · Answer

something is screwy here

anonymous · Answer

are you sure there in an "n" in this?

anonymous · Answer

no it is right, i am screwy.  hold on and let me see what we can do

anonymous · Answer

lord it would help if i knew how to differentiate
sorry

anonymous · Answer

$$f(x)=\ln(\frac{x}{n}+1)$$
$$f'(x)=\frac{1}{x+n}$$ i forgot the damned chain rule!!

anonymous · Answer

$$f'(0)=\frac{1}{n}$$ so the expansion starts with 
$$\frac{x}{n}$$

anonymous · Answer

$$f''(x)=-\frac{1}{(x+n)^2}$$
$$f''(0)=-\frac{1}{n^2}$$ so second term of the expansion is 
$$-\frac{x^2}{2n^2}$$

anonymous · Answer

you got this or should i keep going?

anonymous · Answer

beautiful. it all makes sense now. thank you so very much!!