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 )

Question

11-06-2021
Mathematics

contestada

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 )

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MrRoyal MrRoyal · Answer 1 · 2021-06-14T04:33:35+02:00

Answer:

[tex]\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})[/tex]

Step-by-step explanation:

Given

[tex]\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)[/tex]

Required

The sum

Using the laws of logarithm, we have:

[tex]\log(a) + \log(b) = \log(ab)[/tex]

Take the first two terms of the series

[tex]\log(\frac{1}{2}) + \log(\frac{2}{3}) = \log(\frac{1}{2} * \frac{2}{3})[/tex]

[tex]\log(\frac{1}{2}) + \log(\frac{2}{3}) = \log(\frac{1}{3})[/tex]

Include the third term

[tex]\log(\frac{1}{2}) + \log(\frac{2}{3})+ \log(\frac{3}{4}) = \log(\frac{1}{3} * \frac{3}{4})[/tex]

[tex]\log(\frac{1}{2}) + \log(\frac{2}{3})+ \log(\frac{3}{4}) = \log(\frac{1}{4})[/tex]

Include the fourth term

[tex]\log(\frac{1}{2}) + \log(\frac{2}{3})+ \log(\frac{3}{4}) + \log(\frac{4}{5}) = \log(\frac{1}{4} *\frac{4}{5})[/tex]

[tex]\log(\frac{1}{2}) + \log(\frac{2}{3})+ \log(\frac{3}{4}) + \log(\frac{4}{5}) = \log(\frac{1}{5})[/tex]

Notice the following pattern

[tex]\log(\frac{1}{2}) + \log(\frac{2}{3}) = \log(\frac{1}{3})[/tex] ---------------- n =2

[tex]\log(\frac{1}{2}) + \log(\frac{2}{3})+ \log(\frac{3}{4}) = \log(\frac{1}{4})[/tex] -------------- n = 3

[tex]\log(\frac{1}{2}) + \log(\frac{2}{3})+ \log(\frac{3}{4}) + \log(\frac{4}{5}) = \log(\frac{1}{5})[/tex] ----------- n = 4

So the sum of n series is:

[tex]\log(\frac{1}{2}) + \log(\frac{2}{3})+ ............ + \log(\frac{n}{n+1}) = \log(\frac{1}{n+1})[/tex]

So, the sum of the series is:

[tex]\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)[/tex]

[tex]\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})[/tex]

[tex]\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})[/tex]