Question

How do you use the direct comparison test to determine if `Sigma lnn/(n+1)` from `[2,oo)` is convergent or divergent?

Answer 1

We should first show that `sum_(n=2)^oo1/(n+1)` is divergent through limit comparison.

Letting `a_n=1/n` and `b_n=1/(n+1)` we see that

`lim_(nrarroo)a_n/b_n=lim_(nrarroo)(1/n)/(1/(n+1))=lim_(nrarroo)(n+1)/n=1`

Since `suma_n` is known to be divergent, we see that `sumb_n=sum_(n=2)^oo1/(n+1)` is divergent as well.

For `n>=3`, we see that `lnn>1`. Thus we can say that

`1/(n+1)<=lnn/(n+1)`

Since `1/(n+1)` is already a divergent series, the direct comparison test tells us that `sum_(n=2)^oolnn/(n+1)` is divergent as well.