Why $\lim\limits_{n \to \infty} \frac{ (n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{1}{k+1}$?

Can anyone explain the following equation?

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

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

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

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 $n\to\infty$.

• What does it mean "RHS"? – David Aug 29 '13 at 7:53
• @David: right-hand side (of the equation) – Ron Gordon Aug 29 '13 at 7:53
• Could you elaborate further expansion? – David Aug 29 '13 at 8:28
• @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 $n$. – Ron Gordon Aug 29 '13 at 8:30
• I can not understand the RHS ($\frac{1}{n+1} \frac{1}{1-(1+1/n)^{-(k+1)}}$) – David Aug 29 '13 at 17:16

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

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

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