Find the first four partial sums and the partial sum of the sequence $a _ { n } = \ln ( ( n + 1 ) / ( n + 2 ) )$ .

Name: Find the first four partial sums and the partial sum of the sequence $a _ { n } = \ln ( ( n + 1 ) / ( n + 2 ) )$ .
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Question

Find the first four partial sums and the  partial sum of the sequence $a _ { n } = \ln ( ( n + 1 ) / ( n + 2 ) )$ .

Quizplus · Accepted Answer

The Answer is

S _ { 1 } = \ln \left( \frac { 2 } { 3 } \right) = \ln ( 2 ) - \ln ( 3 )

S _ { 2 } = \ln \left( \frac { 2 } { 3 } \right) + \ln \left( \frac { 3 } { 4 } \right) = \ln ( 2 ) - \ln ( 3 ) + \ln ( 3 ) - \ln ( 4 ) = \ln 2 - \ln ( 4 )

S _ { 3 } = \ln \left( \frac { 2 } { 3 } \right) + \ln \left( \frac { 3 } { 4 } \right) + \ln \left( \frac { 4 } { 5 } \right) = \ln ( 2 ) - \ln ( 3 ) + \ln ( 3 ) - \ln ( 4 ) + \ln ( 4 ) - \ln ( 5 ) = \ln 2 - \ln ( 5 )

S _ { 4 } = \ln \left( \frac { 2 } { 3 } \right) + \ln \left( \frac { 3 } { 4 } \right) + \ln \left( \frac { 4 } { 5 } \right) + \ln \left( \frac { 5 } { 6 } \right) = \ln ( 2 ) - \ln ( 3 ) + \ln ( 3 ) - \ln ( 4 ) + \ln ( 4 ) - \ln ( 5 ) + \ln ( 5 ) - \ln ( 6 ) = \ln 2 - \ln ( 6 )

S _ { n } = \ln \left( \frac { 2 } { n \cdot 2 } \right) = \ln 2 - \ln ( n + 2 )

Find the First Four Partial Sums and the Partial an=ln⁡((n+1)/(n+2))a _ { n } = \ln ( ( n + 1 ) / ( n + 2 ) )an​=ln((n+1)/(n+2))

Find the First Four Partial Sums and the Partial $a _ { n } = \ln ( ( n + 1 ) / ( n + 2 ) )$