In the solution of a problem(How to integrate logarithm and power function?) I encountered the following statement:

For $b \to 0$,and $j \in \mathbb{N}$ we have that $$\binom{b}{j} = (-1)^{j+1} b/j+O(b^2).$$ I checked different books such as "Table of Integrals, Series, and Product", but I could not find it. Can anyone help?

  • $\begingroup$ I'm pretty sure I saw this post yesterday on MSE and it has been answered too. Can someone link this question to its duplicate? Thanks. $\endgroup$ – Prasun Biswas Oct 6 '17 at 19:10
  • $\begingroup$ The same user asked a very similar question yesterday. It was closed. $\endgroup$ – davidlowryduda Oct 6 '17 at 19:40

Write $$ {b \choose j} = \frac{b(b-1)(b-2)\cdots(b-j+1)}{j!}$$ and think as follows. For small $b$, we have $b^2 \ll b$. So the largest part of the numerator will be the first $b$ multiplied by all the following constant terms, which is $b (j-1)! (-1)^{j-1}$. Thus the leading term of the numerator contributes $$ b(-1)^{j-1} \frac{(j-1)!}{j!} = (-1)^{j+1} \frac{b}{j}.$$ The second term in the numerator is $b^2$ multiplied by the second symmetric polynomial of the integers from $-1$ to $-(j-1)$, and so on. In total, the error will look like $O_j(b^2 + b^3 + \cdots + b^{j}) = O_j(b^2)$, where I use heavily that the implicit constant depends on $j$ and that large powers of $b$ decrease very rapidly.

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    $\begingroup$ There's quite a bit of handwaving, here, but most will agree that it could be made rigorous by somebody less lazy (and most won't volunteer). $\endgroup$ – Professor Vector Oct 6 '17 at 20:04

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