How do you use the Nth term test on the infinite series #sum_(n=1)^ooln((2n+1)/(n+1))# ?

Question

How do you use the Nth term test on the infinite series  #sum_(n=1)^ooln((2n+1)/(n+1))# ?

Emilia Aguilar · Accepted Answer

By the nth term test (Divergence Test), we can conclude that the posted series diverges. Recall: Divergence Test
If #lim_{n to infty}a_n ne 0#, then #sum_{n=1}^{infty}a_n# diverges. Let us evaluate the limit.
#lim_{n to infty}ln({2n+1}/{n+1})#
by squeezing the limit inside the log,
#=ln(lim_{n to infty}{2n+1}/{n+1})#
by dividing the numerator and the denominator by #n#,
#=ln(lim_{n to infty}{2n+1}/{n+1}cdot{1/n}/{1/n})
=ln(lim_{n to infty}{2+1/n}/{1+1/n})#
since #1/n to 0#, we have
#=ln2ne 0# By Divergence Test, we may conclude that 
#sum_{n=1}^{infty}ln({2n+1}/{n+1})# diverges. Caution: This test does not detect all divergent series; for example, the harmonic series #sum_{n=1}^{infty}1/n# diverges even though #lim_{n to infty}1/n=0#.

Samantha Adkison · Answer

undefined To use the Nth term test on the infinite series $\sum_{n=1}^\infty \ln\left(\frac{2n+1}{n+1}\right)$, you need to examine the limit as $n$ approaches infinity of the $n$th term of the series. This is done by taking the limit of the general term of the series, $\ln\left(\frac{2n+1}{n+1}\right)$, as $n$ tends to infinity. If this limit is not zero, then the series diverges. If the limit is zero, the test is inconclusive, and other tests may be needed to determine convergence or divergence.