How do you determine if the series #ln(1/2) + ln(1/3) + ln(3/4) + ... +ln[k/(k + 1)] + ....# converges?

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Dec 2, 2017

First of all I think there is a little mistake, the second term would be:

#ln(2/3)# instead of #ln(1/3)#.

So it would be:

#sum_(k=1)^(+oo)ln(k/(k+1))#.

We can use an important logarithmic rule that says:

#ln(a/b)=lna-lnb#.

So:

#sum_(k=1)^(+oo)ln(k/(k+1))=ln(1/2)+ln(2/3)+ln(3/4)+...=#

#=ln1-ln2+ln2-ln3+ln3-ln4+...+lnk-ln(k+1)+...#.

The first term is zero, all the other will cancel each other, but there is always another that survive. The bigger becomes #k# the bigger becomes #ln(k+1)#, the smaller becomes #-ln(k+1)#. So the series will diverge to #-oo#.

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