How to find the limit of
$\lim_{n\rightarrow \infty}n^{2}((1+\frac{p}{n})^{q}-(1+\frac{q}{n})^{p}), (p,q,n \in N)$

I found that it equals to $\lim_{n\rightarrow\infty}(\frac{p^{q}*n^{max(p,q)}}{n^{q-2}}-\frac{q^{p}*n^{max(p,q)}}{n^{p-2}})$
but this look's like wrong statement, because I couldn't find anything simple from it.

What kind of method would work for this one?

  • 1
    $\begingroup$ Hint: What happens when you expand everything with the binomial theorem? What constant term (no $n$) are you left with? $\endgroup$ – Calvin Lin Jan 5 '15 at 10:41
  • 1
    $\begingroup$ hmm.. I think it's $\frac{p^{2}q(q-1)}{2} - \frac{q^{2}p(p-1)}{2} = \frac{pq(p(q-1)-q(p-1))}{2} = \frac{pq(pq-p-pq+q)}{2} = \frac{pq(pq-p-pq+q)}{2} = \frac{pq(q-p)}{2}$ $\endgroup$ – shcolf Jan 5 '15 at 10:45

Note that: $$\left(1+\frac{p}{n}\right)^q = 1 + \binom{q}{1}\frac{p}{n} + \binom{q}{2}\frac{p^2}{n^2} + O\left(\frac{1}{n^3}\right)$$ $$\left(1+\frac{q}{n}\right)^p = 1 + \binom{p}{1}\frac{q}{n} + \binom{p}{2}\frac{q^2}{n^2} + O\left(\frac{1}{n^3}\right)$$

Subtracting and multiplying by $n^2$: $$a_n = \left[\binom{q}{2}p^2-\binom{p}{2}q^2\right] + O\left(\frac{1}{n}\right)$$ where $a_n$ is the term in the limit. Taking $n\rightarrow\infty$, $O(n^{-1})$ terms become zero, so the result is the terms in the square brackets.


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