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May 8, 2011

Interesting identities through partial fractions

Filed under: mathematics — ckrao @ 5:14 am

While working on my previous mathematical post on fractions, I stumbled across the following nice identities:

$\displaystyle \frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$

$\displaystyle \frac{1}{x(x+1)^2} = \frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2}$

$\displaystyle \frac{1}{x(x+1)^3} = \frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2} - \frac{1}{(x+1)^3}$

In general,

$\displaystyle \frac{1}{x(x+1)^n} = \frac{1}{x} - \sum_{i=1}^n\frac{1}{(x+1)^i}.$

We can go further and raise both x and (x+1) to powers:

$\displaystyle \frac{1}{x^2(x+1)^2} = -\frac{2}{x} + \frac{1}{x^2} + \frac{2}{x+1} + \frac{1}{(x+1)^2}$

$\displaystyle \frac{1}{x^2(x+1)^3} = -\frac{3}{x} + \frac{1}{x^2} + \frac{3}{x+1} + \frac{2}{(x+1)^2} + \frac{1}{(x+1)^3}$

$\displaystyle \frac{1}{x^2(x+1)^n} = -\frac{n}{x} + \frac{1}{x^2} + \sum_{i=1}^n \frac{n+1-i}{(x+1)^i}$

Next,

$\displaystyle \frac{1}{x^3(x+1)^3} = \frac{6}{x} - \frac{3}{x^2} + \frac{1}{x^3} - \frac{6}{x+1} - \frac{3}{(x+1)^2} - \frac{1}{(x+1)^3}$

$\displaystyle \frac{1}{x^3(x+1)^4} = \frac{10}{x} - \frac{4}{x^2} + \frac{1}{x^3} - \frac{10}{x+1} - \frac{6}{(x+1)^2} - \frac{3}{(x+1)^3} - \frac{1}{(x+1)^4}$

This suggests a connection with binomial coefficients. We conjecture

$\displaystyle \frac{1}{x^m(x+1)^n} = \sum_{i=1}^m \frac{A_i}{x^i} + \sum_{i=1}^m \frac{B_i}{(x+1)^i},$

where $\displaystyle A_i = (-1)^{m-i}\binom{m+n-1-i}{n-1}, B_i = (-1)^m\binom{m+n-1-i}{m-1}$.

To prove this, we use the technique mentioned in my previous fractions post. To find the $A_i$ we multiply both sides by $x^i$ and move terms with powers of x in the denominator to the other side:

$\begin{array}{rcl} \frac{1}{x^{m-i}(x+1)^n} &=& A_1x^{i-1} + A_2x^{i-2} + ... + A_i + \frac{A_{i+1}}{x} + ... + \frac{A_m}{x^{m-i}} + \sum_{j=1}^n \frac{B_jx^i}{(x+1)^j}\\ \Rightarrow A_1 x^{i-1} + A_2 x^{i-2} + ... + A_i + \sum_{j=1}^n\frac{B_jx^i}{(x+1)^j} &=& \frac{1}{x^{m-i}(x+1)^n} -\left(\frac{A_{i+1}}{x} + ... + \frac{A_m}{x^{m-i}} \right)\\ &=& \frac{1-p(x)(x+1)^n}{x^{m-i}(x+1)^n},\end{array}$

where p(x) is a polynomial. Taking the limit of both sides as $x \rightarrow 0$ requires (m-i) applications of l’Hôpital’s rule on the right side (using the fact that $1-p(x)(x+1)^n$ has root x=0 with multiplicity m-i). We end up with

$\begin{array}{lcl}\displaystyle A_i &=& \frac{1}{(m-i)!}\lim_{x \rightarrow 0}\left(\frac{d}{dx}\right)^{m-i} \frac{1}{(x+1)^n}\\ &=& \frac{1}{(m-i)!}\lim_{x \rightarrow 0} (-n)(-n-1)...(-n-(m-i) + 1)(x+1)^{-n-(m-i)}\\&=& (-1)^{m-i}\binom{m+n-1-i}{n-1}.\end{array}$

Similarly,

$\begin{array}{lcl}\displaystyle B_i &=&\frac{1}{(n-i)!}\lim_{x \rightarrow -1}\left(\frac{d}{dx}\right)^{n-i} \frac{1}{x^m}\\&=&\frac{1}{(n-i)!}\lim_{x \rightarrow -1} (-m)(-m-1)...(-m-(n-i)+1)x^{-m-(n-i)}\\&=& (-1)^{n-i-m-n+i}\binom{m+n-i-1}{n-i}\\&=&(-1)^m\binom{m+n-1-i}{m-1},\end{array}$

as was to be shown.