Question

Find the sum of this series \displaystyle \log\left(\frac{1}{2}\right)+\log\left(\frac{2}{3}\right)+\log\left(\frac{3}{4}\right)+\log\left(\frac{4}{5}\right)+...+\log\left(\frac{98}{99}\right) +\log\left(\frac{99}{100}\right)log( 2 1 ​ )+log( 3 2 ​ )+log( 4 3 ​ )+log( 5 4 ​ )+...+log( 99 98 ​ )+log( 100 99 ​ )

Answer 1

`\log\left((1)/(2)\right)+\log\left((2)/(3)\right)+\log\left((3)/(4)\right)+\log\left((4)/(5)\right)+...+\log\left((98)/(99)\right) +\log\left((99)/(100)\right) = \log((1)/(100))`

Explanation:

Given

`\log\left((1)/(2)\right)+\log\left((2)/(3)\right)+\log\left((3)/(4)\right)+\log\left((4)/(5)\right)+...+\log\left((98)/(99)\right) +\log\left((99)/(100)\right)`

Required

The sum

Using the laws of logarithm, we have:

`\log(a) + \log(b) = \log(ab)`

Take the first two terms of the series

`\log((1)/(2)) + \log((2)/(3)) = \log((1)/(2) * (2)/(3))`

`\log((1)/(2)) + \log((2)/(3)) = \log((1)/(3))`

Include the third term

`\log((1)/(2)) + \log((2)/(3))+ \log((3)/(4)) = \log((1)/(3) * (3)/(4))`

`\log((1)/(2)) + \log((2)/(3))+ \log((3)/(4)) = \log((1)/(4))`

Include the fourth term

`\log((1)/(2)) + \log((2)/(3))+ \log((3)/(4)) + \log((4)/(5)) = \log((1)/(4) *(4)/(5))`

`\log((1)/(2)) + \log((2)/(3))+ \log((3)/(4)) + \log((4)/(5)) = \log((1)/(5))`

Notice the following pattern

`\log((1)/(2)) + \log((2)/(3)) = \log((1)/(3))` ---------------- n =2

`\log((1)/(2)) + \log((2)/(3))+ \log((3)/(4)) = \log((1)/(4))` -------------- n = 3

`\log((1)/(2)) + \log((2)/(3))+ \log((3)/(4)) + \log((4)/(5)) = \log((1)/(5))` ----------- n = 4

So the sum of n series is:

`\log((1)/(2)) + \log((2)/(3))+ ............ + \log((n)/(n+1)) = \log((1)/(n+1))`

So, the sum of the series is:

`\log\left((1)/(2)\right)+\log\left((2)/(3)\right)+\log\left((3)/(4)\right)+\log\left((4)/(5)\right)+...+\log\left((98)/(99)\right) +\log\left((99)/(100)\right)`

`\log\left((1)/(2)\right)+\log\left((2)/(3)\right)+\log\left((3)/(4)\right)+\log\left((4)/(5)\right)+...+\log\left((98)/(99)\right) +\log\left((99)/(100)\right) = \log((1)/(99+1))`

`\log\left((1)/(2)\right)+\log\left((2)/(3)\right)+\log\left((3)/(4)\right)+\log\left((4)/(5)\right)+...+\log\left((98)/(99)\right) +\log\left((99)/(100)\right) = \log((1)/(100))`