ให้ $A_n=\sum^n_{k=1}\frac{(-1)^{k+1}}{k(k+1)}\binom{n}{k}$
$\begin{array}{cl}&{ได้ว่า}\;A_{n+1}-A_n\\
=&\sum^{n+1}_{k=1}\frac{(-1)^{k+1}}{k(k+1)}\binom{n+1}{k}-\sum^n_{k=1}\frac{(-1)^{k+1}}{k(k+1)}\binom{n}{k}\\
=&\frac{(-1)^{n+2}}{(n+1)(n+2)}+\sum^{n}_{k=1}\frac{(-1)^{k+1}}{k(k+1)}\cdot\frac{n+1}{n+1-k}\binom{n}{k}-\sum^n_{k=1}\frac{(-1)^{k+1}}{k(k+1)}\binom{n}{k}\\
=&\frac{(-1)^{n+2}}{(n+1)(n+2)}+\sum^{n}_{k=1}\frac{(-1)^{k+1}}{k(k+1)}\cdot\frac{k}{n+1-k}\binom{n}{k}\\
=&\frac{(-1)^{n+2}}{(n+1)(n+2)}+\sum^{n}_{k=1}\frac{(-1)^{k+1}}{(k+1)(n+1-k)}\binom{n}{k}\\
=&\frac{(-1)^{n+2}}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}\sum^{n}_{k=1}{(-1)^{k+1}}\binom{n+2}{k+1}\\
=&\frac{(-1)^{n+2}}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}\sum^{n+1}_{k=2}{(-1)^{k}}\binom{n+2}{k}\\
=&\frac{(-1)^{n+2}}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}[(\sum^{n+2}_{k=0}{(-1)^{k}}\binom{n+2}{k})-(1-(n+2)+(-1)^{n+2})]\\
=&\frac{(-1)^{n+2}}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}[0-(1-(n+2)+(-1)^{n+2})]\\
=&\frac{1}{n+2}\end{array}$
$\begin{array}{cl}&{ดังนั้น}\;A_n-A_1\\
=&(A_n-A_{n-1})+(A_{n-1}-A_{n-2})+\cdots+(A_2-A_1)\\
=&\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n+1}\end{array}$
$\begin{array}{cl}&\therefore A_n\\
=&A_1+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n+1}\\
=&\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n+1}\end{array}$
${จึงสรุปได้ว่า}\frac{1}{1\cdot 2}\binom{n}{1} - \frac{1}{2\cdot 3}\binom{n}{2}+\frac{1}{3\cdot 4}\binom{n}{3}-...+\frac{(-1)^{n+1}}{n\cdot (n+1)}\binom{n}{n}=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n+1}$