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edited Dec 25, 2015 at 17:08

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This is a specialization at $x = n$ of the rational function identity $$ \frac{1}{x(x+1)\cdots(x+m)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{x+k}, $$ and this identity could be proved either by partial fractions with unknown coefficients and limits to find the coefficients (see my comment on Todd Trimble's answer) or by induction on $m$. For the inductive step, divide both sides of the above identity by $x+m+1$: $$ \frac{1}{x(x+1)\cdots(x+m)(x+m+1)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{(x+k)(x+m+1)}. $$ Split up the right side by partial fractions: to make it $$ \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{(x+k)(x+m+1)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{m+1-k}\left(\frac{1}{x+k} - \frac{1}{x+m+1}\right). $$$$ \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{m+1-k}\left(\frac{1}{x+k} - \frac{1}{x+m+1}\right). $$ Break up the right sidethis into a sum of two terms: $$ \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{1}{m+1-k}\frac{(-1)^k}{x+k} - \frac{1}{m!}\left(\sum_{k=0}^m\binom{m}{k}\frac{(-1)^k}{m+1-k}\right)\frac{1}{x+m+1}. $$ Since $\binom{m}{k}/(m+1-k) = \binom{m+1}{k}/(m+1)$, the difference is $$ \frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\left(\sum_{k=0}^m\binom{m+1}{k}(-1)^k\right)\frac{1}{x+m+1}. $$ The sum inside parentheses is $(1-1)^{m+1} - (-1)^{m+1} = (-1)^m$, so finally we have $$ \frac{1}{x(x+1)\cdots(x+m)(x+m+1)} = \frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\frac{(-1)^m}{x+m+1}. $$ Since $-(-1)^m = (-1)^{m+1}$, we're done.

This is a specialization at $x = n$ of the rational function identity $$ \frac{1}{x(x+1)\cdots(x+m)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{x+k}, $$ and this identity could be proved either by partial fractions with unknown coefficients and limits to find the coefficients (see my comment on Todd Trimble's answer) or by induction on $m$. For the inductive step, divide both sides of the above identity by $x+m+1$: $$ \frac{1}{x(x+1)\cdots(x+m)(x+m+1)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{(x+k)(x+m+1)}. $$ Split up the right side by partial fractions: $$ \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{(x+k)(x+m+1)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{m+1-k}\left(\frac{1}{x+k} - \frac{1}{x+m+1}\right). $$ Break up the right side into a sum of two terms: $$ \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{1}{m+1-k}\frac{(-1)^k}{x+k} - \frac{1}{m!}\left(\sum_{k=0}^m\binom{m}{k}\frac{(-1)^k}{m+1-k}\right)\frac{1}{x+m+1}. $$ Since $\binom{m}{k}/(m+1-k) = \binom{m+1}{k}/(m+1)$, the difference is $$ \frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\left(\sum_{k=0}^m\binom{m+1}{k}(-1)^k\right)\frac{1}{x+m+1}. $$ The sum inside parentheses is $(1-1)^{m+1} - (-1)^{m+1} = (-1)^m$, so finally we have $$ \frac{1}{x(x+1)\cdots(x+m)(x+m+1)} = \frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\frac{(-1)^m}{x+m+1}. $$ Since $-(-1)^m = (-1)^{m+1}$, we're done.

This is a specialization at $x = n$ of the rational function identity $$ \frac{1}{x(x+1)\cdots(x+m)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{x+k}, $$ and this identity could be proved either by partial fractions with unknown coefficients and limits to find the coefficients (see my comment on Todd Trimble's answer) or by induction on $m$. For the inductive step, divide both sides of the above identity by $x+m+1$: $$ \frac{1}{x(x+1)\cdots(x+m)(x+m+1)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{(x+k)(x+m+1)}. $$ Split up the right side by partial fractions to make it $$ \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{m+1-k}\left(\frac{1}{x+k} - \frac{1}{x+m+1}\right). $$ Break this into a sum of two terms: $$ \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{1}{m+1-k}\frac{(-1)^k}{x+k} - \frac{1}{m!}\left(\sum_{k=0}^m\binom{m}{k}\frac{(-1)^k}{m+1-k}\right)\frac{1}{x+m+1}. $$ Since $\binom{m}{k}/(m+1-k) = \binom{m+1}{k}/(m+1)$, the difference is $$ \frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\left(\sum_{k=0}^m\binom{m+1}{k}(-1)^k\right)\frac{1}{x+m+1}. $$ The sum inside parentheses is $(1-1)^{m+1} - (-1)^{m+1} = (-1)^m$, so finally we have $$ \frac{1}{x(x+1)\cdots(x+m)(x+m+1)} = \frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\frac{(-1)^m}{x+m+1}. $$ Since $-(-1)^m = (-1)^{m+1}$, we're done.

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created Jan 11, 2015 at 4:43

KConrad

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This is a specialization at $x = n$ of the rational function identity $$ \frac{1}{x(x+1)\cdots(x+m)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{x+k}, $$ and this identity could be proved either by partial fractions with unknown coefficients and limits to find the coefficients (see my comment on Todd Trimble's answer) or by induction on $m$. For the inductive step, divide both sides of the above identity by $x+m+1$: $$ \frac{1}{x(x+1)\cdots(x+m)(x+m+1)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{(x+k)(x+m+1)}. $$ Split up the right side by partial fractions: $$ \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{(x+k)(x+m+1)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{m+1-k}\left(\frac{1}{x+k} - \frac{1}{x+m+1}\right). $$ Break up the right side into a sum of two terms: $$ \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{1}{m+1-k}\frac{(-1)^k}{x+k} - \frac{1}{m!}\left(\sum_{k=0}^m\binom{m}{k}\frac{(-1)^k}{m+1-k}\right)\frac{1}{x+m+1}. $$ Since $\binom{m}{k}/(m+1-k) = \binom{m+1}{k}/(m+1)$, the difference is $$ \frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\left(\sum_{k=0}^m\binom{m+1}{k}(-1)^k\right)\frac{1}{x+m+1}. $$ The sum inside parentheses is $(1-1)^{m+1} - (-1)^{m+1} = (-1)^m$, so finally we have $$ \frac{1}{x(x+1)\cdots(x+m)(x+m+1)} = \frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\frac{(-1)^m}{x+m+1}. $$ Since $-(-1)^m = (-1)^{m+1}$, we're done.