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.


Hint: \begin{align*} \big( 1+\tfrac pn \big)^{n/2p} &= \exp\big( \tfrac n{2p} \log(1+\tfrac pn) \big) \\ &= \exp\big( \tfrac n{2p} (\tfrac pn - \tfrac12\tfrac{p^2}{n^2} + \tfrac13\tfrac{p^3}{n^3}) - \cdots \big) \\ &= \exp\big( \tfrac12 - \tfrac14\tfrac{p}{n} + \tfrac16\tfrac{p^2}{n^2} - \cdots \big) \\ &= e^{1/2} \bigg( 1 + \big( - \tfrac14\tfrac{p}{n} + \tfrac16\tfrac{p^2}{n^2} - \cdots \big) + \tfrac1{2!} \big( - \tfrac14\tfrac{p}{n} + \cdots \big)^2 + \cdots \bigg). \end{align*}

  • $\begingroup$ Very good idea ! In the same spirit, I propose a shortcut which allows to handle the problem slightly faster. The merit is your. Cheers :) $\endgroup$ – Claude Leibovici Jun 27 '14 at 18:50
  • $\begingroup$ 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 $ep^2/3$ $\endgroup$ – user182973 Jun 27 '14 at 21:24
  • $\begingroup$ Yes, although I don't think l'Hopital is even needed: the expression inside the outer parentheses of the original problem just simplifies to $e(\frac{p^2}{3n^2} + O(\frac{p^3}{n^3}))$. $\endgroup$ – 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.

By the way, the result is beautiful !


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