Question

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

Answer 1

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`.