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How do you find the 5-th partial sum of the infinite series #sum_(n=1)^ooln((n+1)/n)# ?

Answer 1

To find the 5th partial sum of the infinite series ( \sum_{n=1}^\infty \ln\left(\frac{n+1}{n}\right) ), we need to sum the first five terms of the series.

The general term of the series is ( \ln\left(\frac{n+1}{n}\right) ).

Substituting ( n = 1, 2, 3, 4, ) and ( 5 ) into the general term, we get:

[ \ln\left(\frac{2}{1}\right) + \ln\left(\frac{3}{2}\right) + \ln\left(\frac{4}{3}\right) + \ln\left(\frac{5}{4}\right) + \ln\left(\frac{6}{5}\right) ]

[ = \ln(2) + \ln\left(\frac{3}{2}\right) + \ln\left(\frac{4}{3}\right) + \ln\left(\frac{5}{4}\right) + \ln\left(\frac{6}{5}\right) ]

[ = \ln\left(2 \times \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \frac{6}{5}\right) ]

[ = \ln(6) ]

Therefore, the 5th partial sum of the series is ( \ln(6) ).

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Answer 2

Ezra Adams

#S_5=sum_{n=1}^5ln({n+1}/n)#

by writing them all out, #=ln(2/1)+ln(3/2)+ln(4/3)+ln(5/4)+ln(6/5)#

by repeatedly applying the log property #lnx+lny=ln(xy)#, #=ln(2/1cdot3/2cdot4/3cdot5/4cdot6/5)#

by cancelling common factors, #=ln6#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.