Why is the following true?
\begin{equation} (1+o(1))n^{r+2}2^{-r}+n_{(r+1)}-n_{(r)}2^{r}(n-r)^{2}2^{-2r}=o(n^{r+2}), \end{equation}
where $n_{(r)}=n(n-1) \cdots (n-r+1)$.
In my opinion, the result should be $o(n^{r+3})$.
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guess the $n^{r+2}$ term cancels out? – SK19 Jan 05 '18 at 08:52
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Ah thank you @SK19 , yes $(1+o(1))n^{r+2}2^{-r}$ and $n_{(r)}(n-r)^{2}2^{-r}$ are asymptoticly the same. I don't know, why I did not see this, but problem solved now. – koala Jan 05 '18 at 08:59
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You can try to write down a detailed answer to your question. On one hand you would exercise explaining mathematical thoughts, on the other hand this question would not stay without an answer :-) – SK19 Jan 06 '18 at 17:15
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Good idea! We have: \begin{equation} (1+o(1))n^{r+2}2^{-r}-n_{(r)}(n-r)^{2}2^{-r}+n_{(r+1)}. \end{equation} We can show, that $(1+o(1))n^{r+2}=n_{(r)}(n-r)^{2}$. We know, that $n^{r+2}$ and $n_{(r)}(n-r)^{2}$ are asymptoticly the same, since $n_{(r)}(n-r)^{2} \leq n^{r+2}$. Therefore, there is a sequence $c_{n}$ such that $n^{r+2}c_{n}= n_{(r)}(n-r)^{2}$ for $n \to \infty$. If we write $1+o(1)$ instead of $c_{n}$ we see, that there is only $n_{(r+1)}$ left, which is $o(n^{r+2})$. – koala Jan 06 '18 at 20:01
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I meant answer in the sense of "Do you really want to answer your own question?" ;) – SK19 Jan 07 '18 at 10:00