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

Answer:

$\log\left(\frac{1}{2}\right)+\log\left(\frac{2}{3}\right)+\log\left(\frac{3}{4}\right)+\log\left(\frac{4}{5}\right)+...+\log\left(\frac{98}{99}\right) +\log\left(\frac{99}{100}\right) = \log(\frac{1}{100})$

Step-by-step explanation:

Given

$\log\left(\frac{1}{2}\right)+\log\left(\frac{2}{3}\right)+\log\left(\frac{3}{4}\right)+\log\left(\frac{4}{5}\right)+...+\log\left(\frac{98}{99}\right) +\log\left(\frac{99}{100}\right)$

Required

The sum

Using the laws of logarithm, we have:

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

Take the first two terms of the series

$\log(\frac{1}{2}) + \log(\frac{2}{3}) = \log(\frac{1}{2} * \frac{2}{3})$

$\log(\frac{1}{2}) + \log(\frac{2}{3}) = \log(\frac{1}{3})$

Include the third term

$\log(\frac{1}{2}) + \log(\frac{2}{3})+ \log(\frac{3}{4}) = \log(\frac{1}{3} * \frac{3}{4})$

$\log(\frac{1}{2}) + \log(\frac{2}{3})+ \log(\frac{3}{4}) = \log(\frac{1}{4})$

Include the fourth term

$\log(\frac{1}{2}) + \log(\frac{2}{3})+ \log(\frac{3}{4}) + \log(\frac{4}{5}) = \log(\frac{1}{4} *\frac{4}{5})$

$\log(\frac{1}{2}) + \log(\frac{2}{3})+ \log(\frac{3}{4}) + \log(\frac{4}{5}) = \log(\frac{1}{5})$

Notice the following pattern

$\log(\frac{1}{2}) + \log(\frac{2}{3}) = \log(\frac{1}{3})$ ---------------- n =2

$\log(\frac{1}{2}) + \log(\frac{2}{3})+ \log(\frac{3}{4}) = \log(\frac{1}{4})$ -------------- n = 3

$\log(\frac{1}{2}) + \log(\frac{2}{3})+ \log(\frac{3}{4}) + \log(\frac{4}{5}) = \log(\frac{1}{5})$ ----------- n = 4

So the sum of n series is:

$\log(\frac{1}{2}) + \log(\frac{2}{3})+ ............ + \log(\frac{n}{n+1}) = \log(\frac{1}{n+1})$

So, the sum of the series is:

$\log\left(\frac{1}{2}\right)+\log\left(\frac{2}{3}\right)+\log\left(\frac{3}{4}\right)+\log\left(\frac{4}{5}\right)+...+\log\left(\frac{98}{99}\right) +\log\left(\frac{99}{100}\right)$

$\log\left(\frac{1}{2}\right)+\log\left(\frac{2}{3}\right)+\log\left(\frac{3}{4}\right)+\log\left(\frac{4}{5}\right)+...+\log\left(\frac{98}{99}\right) +\log\left(\frac{99}{100}\right) = \log(\frac{1}{99+1})$

$\log\left(\frac{1}{2}\right)+\log\left(\frac{2}{3}\right)+\log\left(\frac{3}{4}\right)+\log\left(\frac{4}{5}\right)+...+\log\left(\frac{98}{99}\right) +\log\left(\frac{99}{100}\right) = \log(\frac{1}{100})$