# Find $\lim_{n \rightarrow \infty} n^2 \left(\left(\frac{1+p/n}{1-p/n}\right)^{n/2p} - e\right)$

Let $p$ be an arbitrary real number. I wish to compute the following limit

$$\lim_{n \rightarrow \infty} n^2 \left(\left(\frac{1+p/n}{1-p/n}\right)^{n/2p} - e\right)$$

I've tried to express $((1+p/n)/(1-p/n))^{n/2p}$ as $(1 + x/k)^k$ for some $x,k$, but I can't seem to be able to do this. Not sure else how to proceed.

• Aha. So the trick is to use the above for the numerator, and similarly for the denominator, then apply L'Hopital's rule. The answer should be $ep^2/3$ – user182973 Jun 27 '14 at 21:24
• Yes, although I don't think l'Hopital is even needed: the expression inside the outer parentheses of the original problem just simplifies to $e(\frac{p^2}{3n^2} + O(\frac{p^3}{n^3}))$. – Greg Martin Jun 27 '14 at 22:58
Exactly in the same spirit of Greg Martin's answer, directly use (it will be faster) $$\log \left(\frac{1+x}{1-x}\right)=2 x+\frac{2 x^3}{3}+\frac{2 x^5}{5}+O\left(x^6\right)$$ Replace $x$ by $\frac{p}{n}$ and continue as Greg Martin suggested.