Solution verification of the convergence of $\sum_{n=1}^\infty \frac{(-1)^n}{x+n}$

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.

• First, uniform convergence and conditional convergence are not the same. Second, yes you are meant to consider $x\in(0,\infty)$. – TSF Aug 14 '20 at 23:40
• @TSF Many thanks. I decided to edit the question and ask as if it was truly asking for the conditional convergence. – tau20 Aug 14 '20 at 23:44
• When you do the comparison test, your inequality is wrong. $\frac{1}{n} > \frac{1}{n+x}$ for all $x\in(0,\infty)$. – TSF Aug 14 '20 at 23:57

Yes, the Leibniz test shows the conditional convergence. However, your use of the comparison test is wrong since you start with,

$$\frac{1}{n} < \frac{1}{n+x}$$

which is not true for $$x\in(0,\infty)$$. Think of $$n=1$$ with $$x=1$$,

$$\frac{1}{n} = 1 > \frac{1}{2} = \frac{1}{n+x}$$

You need to fix $$x\in(0,\infty)$$ and then lower bound $$\frac{1}{x+n}$$ by a term from the harmonic series, but $$\frac{1}{n}$$ is not the one you want.

• I see, many thanks. I will try to figure it out. Just to be clear, was the abs. divergence proven correctly with the use of Raabe criteria? – tau20 Aug 15 '20 at 0:07
• Never heard of it. – TSF Aug 15 '20 at 0:14
• archive.lib.msu.edu/crcmath/math/math/r/r006.htm – zkutch Aug 15 '20 at 0:51

Unfortunately, your use of the Raabe criterion was also incorrect. The Raabe test involves the computation of the limit $$\lim_{n\to\infty} n \left(\frac{a_n}{a_{n+1}} - 1\right) = \lim_{n\to\infty} n \left(\frac{x+n+1}{x+n} - 1\right),$$ but in your answer, you dropped the factor of $$n$$. With this corrected, the Raabe criterion is inconclusive: $$\lim_{n\to\infty} n \left(\frac{x+n+1}{x+n} - 1\right) = \lim_{n\to\infty} \frac{n}{x+n}= 1.$$

• I see, many thanks. – tau20 Aug 15 '20 at 17:55