Nth Term Test for Divergence of an Infinite Series

Key Questions

• 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.
${\lim}_{n \to \infty} \ln \left(\frac{2 n + 1}{n + 1}\right)$
by squeezing the limit inside the log,
$= \ln \left({\lim}_{n \to \infty} \frac{2 n + 1}{n + 1}\right)$
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 $\frac{1}{n} \to 0$, we have
$= \ln 2 \ne 0$

By Divergence Test, we may conclude that
${\sum}_{n = 1}^{\infty} \ln \left(\frac{2 n + 1}{n + 1}\right)$ diverges.

Caution: This test does not detect all divergent series; for example, the harmonic series ${\sum}_{n = 1}^{\infty} \frac{1}{n}$ diverges even though ${\lim}_{n \to \infty} \frac{1}{n} = 0$.

• Nth Term Test (also called Divergence Test)

If ${\lim}_{n \to \infty} | {a}_{n} | \ne 0$, then ${\sum}_{n = 1}^{\infty} {a}_{n}$ diverges.