# Order of the maximum term of (a +b)^(-m)

I would like, if possible, to obtain a proof of the theorem below.

"Being given real numbers a and b, with |a|>|b|, and m is a positive integer, the order p, which occupies the maximum term (in absolute value) the development of power (a+b)^(-m) , according to decreasing powers of a, is given by:

p = 1 + integer part of [|b|(m-1)/(|a|-|b|)]

When the expression |b|(m-1)/ (|a|-|b|) is an integer, then there are maximum two terms (in absolute value): they are of order p and p-1. "

Paulo Argolo

Rio de Janeiro, Brazil

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The expansion is

$\sum_{p=0}^{\infty} \binom{-m}{p} a^{-m-p} b^p$

The ratio between the absolute values of the terms corresponding to $p+1$ and to $p$ is

$\frac{m+p}{p+1} \cdot \frac{|b|}{|a|}$

The $p+1$st term is larger than or equal to the $p$th term if

$m+p \geq (p+1)\frac{|a|}{|b|}$

or

$p \left(1 - \frac{|a|}{|b|}\right) \geq \frac{|a|}{|b|} - m$

Since $|a| < |b|$, this is the same as

$p \leq \frac{|a|/|b|-m}{1-|a|/|b|} = \frac{|a|-m|b|}{|b|-|a|}$

or

$p + 1 \leq \frac{(1-m)|b|}{|b|-|a|} = \frac{(m-1)|b|}{|a|-|b|}$

So the $p$th term is maximal if

$p \leq \frac{(m-1)|b|}{|a|-|b|}$

and

$p+1 \geq \frac{(m-1)|b|}{|a|-|b|}$

If the fraction is an integer $N$, then both $p = N$ and $p = N - 1$ are solutions. Otherwise, $p$ is the floor of the fraction, and I'm off by one wrt your solution.

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