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

Question

Isabella Ackerman · Accepted Answer

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.

Christopher Agresta · Answer

undefined To determine the convergence or divergence of the series $ \sum_{n=2}^{\infty} \frac{\ln n}{n+1} $ using the direct comparison test, we compare it to a known series.

We notice that $ \frac{\ln n}{n+1} $ is similar to $ \frac{1}{n} $ as $ n $ approaches infinity, since the natural logarithm grows slower than any power of $ n $.

Now, we compare the given series to the harmonic series $ \sum_{n=1}^{\infty} \frac{1}{n} $, which is known to be divergent. Specifically, we compare $ \frac{\ln n}{n+1} $ to $ \frac{1}{n} $ for $ n \geq 2 $.

Since $ \frac{\ln n}{n+1} < \frac{1}{n} $ for all $ n \geq 2 $, and the harmonic series $ \sum_{n=1}^{\infty} \frac{1}{n} $ diverges, by the direct comparison test, the series $ \sum_{n=2}^{\infty} \frac{\ln n}{n+1} $ also diverges.