1
$\begingroup$

Can anyone explain the following equation?

$$\lim_{n \to \infty} \frac{ (n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{1}{k+1}$$

$\endgroup$
  • 1
    $\begingroup$ Let $a=n+1$, $b=n$. Use $a^m-b^m=(a-b)(a^{m-1}+a^{m-2}{b}+\cdots+b^{m-1})$. $\endgroup$ – André Nicolas Aug 29 '13 at 8:03
5
$\begingroup$

$$\frac{(n+1)^{k}}{(n+1)^{k+1}-n^{k+1}} = \frac{1}{n+1} \frac{1}{1-(1+1/n)^{-(k+1)}}$$

Taylor expand in the denominator in the RHS:

$$\frac{(n+1)^{k}}{(n+1)^{k+1}-n^{k+1}} \approx \frac{1}{n+1} \frac{1}{1-(1-(k+1)/n)}=\frac{n}{n+1} \frac{1}{k+1}$$

The result follows from taking the limit as $n\to\infty$.

$\endgroup$
  • $\begingroup$ What does it mean "RHS"? $\endgroup$ – David Aug 29 '13 at 7:53
  • $\begingroup$ @David: right-hand side (of the equation) $\endgroup$ – Ron Gordon Aug 29 '13 at 7:53
  • $\begingroup$ Could you elaborate further expansion? $\endgroup$ – David Aug 29 '13 at 8:28
  • $\begingroup$ @David: $$(1+y)^{-\alpha} \approx 1- \alpha y$$ for small $y$. In our case, $y=1/n$ is small because we are only considering the limit of very large $n$. $\endgroup$ – Ron Gordon Aug 29 '13 at 8:30
  • $\begingroup$ I can not understand the RHS ($ \frac{1}{n+1} \frac{1}{1-(1+1/n)^{-(k+1)}} $) $\endgroup$ – David Aug 29 '13 at 17:16
5
$\begingroup$

Apply L'Hôpital's rule $k$ times to get $$\lim_{n \to \infty} \frac{ k!}{(k+1)!((n+1)-n)}=\frac{1}{k+1}$$

$\endgroup$
  • $\begingroup$ I think you have to say a bit more since $n$ is discrete $\endgroup$ – Belgi Aug 29 '13 at 8:06
1
$\begingroup$

This is equivalent to one of @Ron's steps that $$a_mx^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0\sim a_mx^m$$ while $x\to\infty$. So $$\frac{(n+1)^{k}}{(n+1)^{k+1}-n^{k+1}} \sim \frac{n^k}{(k+1)n^k}= \frac{1}{k+1},~~n\to\infty$$

$\endgroup$

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.