# Interesting limit: $I(n)=\lim_{x \to 0} (e^x-1)^{-n} - \left(\frac{x}{1!}+\frac {x^2}{2!}+\dotsb+\frac{x^n}{n!}\right)^{-n}$

I recently came up with a problem which I think is quite interesting and here it is.

Let $$I(n)=\lim_{x\to 0} \left(\frac {1}{(e^x-1)^n} -\frac {1}{\left(\frac {x}{1!}+\frac {x^2}{2!}+\frac {x^3}{3!}+\cdots +\frac {x^n}{n!}\right)^n}\right)$$ Evaluate $$I(n)$$ in terms of $$n$$.

Now I tried a lot of messing ups with this monster. Tried L'Hospital, Squeeze theorem, Stolz-Cesaro, Binomial theorem and Multinomial theorem, etc. But couldn't reach the general form. On writing out few terms using Wolfy I get $$I(1)=\frac {-1}{2}$$ , $$I(2)=\frac {-1}{3}$$, $$I(3)=\frac {-1}{8}$$, $$I(4)=\frac {-1}{30}$$ and so on

On spending some time on these observations I could conjecture that

$$I(n)=\frac {-n}{(n+1)!}$$

and tried this formula to check whether it was consistent with other values of $$n$$ and it indeed was. So Now I have the general form but no proof to it.

And something which is quite interesting in this question is that the first fraction has a denominator with a function raised to power $$n$$ while the denominator of second fraction is nothing but the Maclaurin expansion of that function with finite terms (here $$n$$ terms) raised to power of $$n$$. So I tried to test it with some other functions and the results impressed me a lot.

1) $$J(n)=\lim_{x\to 0} \left(\frac {1}{(\ln (1+x))^n} -\frac {1}{\left(x-\frac {x^2}{2}+\frac {x^3}{3}-\cdots +\frac {(-1)^{n+1} x^n}{n}\right)^n}\right) =\frac {(-1)^{n+1}n}{n+1}$$

2)$$G(n)=\lim_{x\to 0} \left(\frac {1}{(\arccos x-\frac {\pi}{2})^n} -\frac {1}{\left(\underbrace{\frac {x}{1}+\frac {x^3}{6}+\frac {3x^5}{40}+\frac {5x^7}{112}+\cdots}_{\text {n terms}}\right)^n}\right)$$ Now this limit is more interesting because it equals $$0$$ for all even natural $$n$$ and it doesn't exist for any odd natural $$n$$

Now what I actually want is a proper method to solve such problems. You can take any of top 2 limits ( i.e of $$(e^x-1)$$ or $$\ln (1+x)$$) to demonstrate your method.

Let $a=e^x-1$ and $b=\frac {x}{1!}+\frac {x^2}{2!}+\frac {x^3}{3!}+\cdots +\frac {x^n}{n!}$, so is it easy to see: $a\sim x$ , $b\sim x$ and $b-a\sim -\frac{x^{n+1}}{(n+1)!}$ as $x\to 0.$ $$\frac {1}{(e^x-1)^n} -\frac {1}{\left(\frac {x}{1!}+\frac {x^2}{2!}+\frac {x^3}{3!}+\cdots +\frac {x^n}{n!}\right)^n}$$ $$=\frac{b^n-a^n}{a^nb^n}=\frac{(b-a)(b^{n-1}+b^{n-2}a+\cdots+a^{n-1})}{a^n b^n}$$ $$\sim\frac{\left(-\frac{x^{n+1}}{(n+1)!}\right)(nx^{n-1})}{x^{2n}}\to\frac {-n}{(n+1)!},$$ as $x\to 0$.