Question

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

Answer 1

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`.