1
$\begingroup$

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?

$\endgroup$
  • 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.