This is an exam problem from the analysis 1 course I attend.
Problem: Test the following series for absolute and conditional convergence $$ \sum_{n=1}^\infty \frac{(-1)^n}{x+n} $$ on the interval $(0,+\infty)$.
Solution:
Testing absokute convergence:
$$\sum_{n=1}^\infty \left| \frac{(-1)^n}{x+n}\right|=\sum_{n=1}^\infty \frac{1}{x+n} $$
1st way to test for absolute convergence using the Raabe criteria:
$$\lim_{n \to \infty}\frac{x+n+1}{x+n}-1=0\ \ \text{,}$$
for all $x \in (0,+\infty)$.
Since $0 < 1$, according to Raabe criteria the series absolutely diverge.
2nd way to test for absolute convergence using the comparison test:
$\text{Since}\ \ \frac {1}{n} < \frac{1}{x+n}\ \ \text{for all }x \in (0,+\infty),\ \text{and since the Harmonic series diverge, whose general term is}\ \ \frac{1}{n},\text {the series absolutely diverge according to the comparison test.}$
Testing conditional convergence(using Leibniz convergence criteria)
1. $$\left|a_n+1\right| < \left|a_n\right|, \frac{1}{x+n+1} < \frac{1}{x+n} $$ We can see that the sequence is monotonically decreasing for all $x \in (0,+\infty)$.
2. $$ \lim_{n \to \infty} \frac{1}{x+n}=0\ \ \text{,}$$
for all $x \in (0,+\infty)$.
And thereby, by the Leibniz criteria, the series converge conditionally.
Is everything alright with the solution ? Feel free to add suggestions if you have any. Thanks.