Can anyone explain the following equation?
$$\lim_{n \to \infty} \frac{ (n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{1}{k+1}$$
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1Let $a=n+1$, $b=n$. Use $a^m-b^m=(a-b)(a^{m-1}+a^{m-2}{b}+\cdots+b^{m-1})$. – André Nicolas Aug 29 '13 at 08:03
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Following @AndréNicolas' comment, $$\lim_{n\to\infty}\frac{(n+1)^m}{(n+1)^{m+1}-n^{m+1}}=\lim_{n\to\infty}\frac{(n+1)^m}{\sum_{k=0}^m(n+1)^kn^{m-k}}=\frac1{1+m}$$ – Matcha Latte Aug 13 '24 at 12:27
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$$\frac{(n+1)^{k}}{(n+1)^{k+1}-n^{k+1}} = \frac{1}{n+1} \frac{1}{1-(1+1/n)^{-(k+1)}}$$
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 $n\to\infty$.
Ron Gordon
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@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 08:30
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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
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Apply L'Hôpital's rule $k$ times to get $$\lim_{n \to \infty} \frac{ k!}{(k+1)!((n+1)-n)}=\frac{1}{k+1}$$
Džuris
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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 $$\frac{(n+1)^{k}}{(n+1)^{k+1}-n^{k+1}} \sim \frac{n^k}{(k+1)n^k}= \frac{1}{k+1},~~n\to\infty$$
Mikasa
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