Please help me calculate the following sum
$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}$
Please help me calculate the following sum
$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}$
HINT: $\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}$
$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}\\=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{99}-\frac{1}{100}\\=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}=0.49$
$\frac{1}{2 \cdot 3} =\frac{1}{2} - \frac{1}{3}$ $\frac{1}{3 \cdot 4}=\frac{1}{3} -\frac{1}{4}$ $\ldots$ $\frac{1}{99 \cdot 100}= \frac{1}{99}-\frac{1}{100}$
So:
$\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{99 \cdot 100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+ \ldots-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}.$
general case:
$\frac{1}{n \cdot (n+1)}=\frac{1}{n}-\frac{1}{n+1}.$