# Question #373e2

The series ${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \frac{n}{{n}^{2} + 1}$ converges by the Alternating Series Test (since its terms alternate in sign and since $\frac{n}{{n}^{2} + 1}$ forms a decreasing sequence that approaches 0 as $n \to \infty$).
However, the corresponding series ${\sum}_{n = 0}^{\infty} | {\left(- 1\right)}^{n} \frac{n}{{n}^{2} + 1} | = {\sum}_{n = 0}^{\infty} \frac{n}{{n}^{2} + 1}$ diverges. This can be seen by noting that $\frac{n}{{n}^{2} + 1} \ge q \frac{1}{2 n} > 0$ for all integers $n \ge q 1$, that ${\sum}_{n = 1}^{\setminus \infty} \frac{1}{2 n}$ diverges (its terms are one-half the terms of the divergent harmonic series ), and then applying the Comparison Test .
The facts in the two paragraphs above mean that the original series ${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \frac{n}{{n}^{2} + 1}$ converges conditionally .