# Which sequence ${a_n}$ does $\sum_{n=1}^\infty a_n$ is conditionally convergent and $\sum_{n=1}^\infty (-1)^n a_n$ converges

Which sequence $${a_n}$$ does $$\sum_{n=1}^\infty a_n$$ is conditionally convergent and $$\sum_{n=1}^\infty (-1)^n a_n$$ converges? I tried with $${a_n}= \frac{\sin(n)}{n}$$ and it seems that it does. But I would like to know if there is another.

• Are you sure you aren't mixing up sequences and series? If $\sum a_n$ converges, $a_n\to 0$; in particular, as a sequence, $a_n$ is absolutely convergent, since $a_n\to 0$ if and only if $|a_n|\to 0$ as well. Commented Jun 15, 2020 at 3:00
• Yes, im sorry.I already corrected it. Commented Jun 15, 2020 at 3:12
• do you need $\sum_{n=1}^\infty |a_n| = \infty$? Commented Jun 15, 2020 at 3:20
• @user251257 Yes :D Commented Jun 15, 2020 at 3:21

Take your favorite conditionally convergent series $$\sum_{n=1}^\infty b_n$$, e.g. $$b_n = (-1)^{n+1} \frac{1}{n}$$. Then define $$a_{2n} = b_n$$ and $$a_{2n - 1} = 0$$. Then, $$\sum_{k=1}^\infty a_k = \sum_{n=1}^\infty b_n \in\mathbb R$$ and $$\sum_{k=1}^\infty |a_k| = \sum_{n=1}^\infty |b_n| = \infty$$ and $$\sum_{k=1}^\infty (-1)^k a_k = \sum_{n=1}^\infty b_n \in\mathbb R.$$

• Thank you a lot. I couldn't have done it without your help Commented Jun 15, 2020 at 3:43

Any alternating series that is convergent has an argument (excluding the $$(-1)^n$$ component) that is conditionally convergent. Examples of some are p-series with $$p\geq1$$ or $$\sum_{n=0}^{\infty}\frac{1}{n!}$$, and there are countless others.

• be careful. The alternating harmonic series converges, the harmonic series does not converge. Commented Jun 15, 2020 at 3:14
• I agree, but he wants to know about conditional convergence @user251257 Commented Jun 15, 2020 at 3:14
• I already tried the p series, but they are not conditionally convergent. Commented Jun 15, 2020 at 3:16
• Why do you say that @JessFlo, all p-series with $p>1$ are absolutely convergent (and all absolutely convergent series are conditionally convergent) and the alternating harmonic series is convergent so they are indeed conditionally convergent. Commented Jun 15, 2020 at 3:18
• @justaguy I need that $\sum_{n=1}^\infty |a_n|$ diverges Commented Jun 15, 2020 at 3:25

Let $$\zeta := e^{2 \pi i/3} = \frac{-1+i\sqrt{3}}{2}$$ be a primitive 3rd root of unity. Then $$\xi :=-\zeta = \frac{1-i\sqrt{3}}{2} = e^{5\pi i/3}$$ is a primitive 6th root of unity, and the partial sums

\begin{align*} \zeta + \zeta^2+&...+\zeta^n \\ \xi +\xi^2 +&...+\xi^n \ \end{align*}

are both bounded for all $$n \geq 0$$ (by the sum of a finite geometric series). Therefore, by Dirichlet's test,

$$\sum_{n=1}^\infty \frac{\zeta^n}{n^s} \text{ and } \sum_{n=1}^\infty \frac{\xi^n}{n^s} =\sum_{n=1}^\infty \frac{(-1)^n\zeta^n}{n^s}$$

are both conditionally convergent for any $$s$$ with $$0 < s \leq 1$$. So $$a_n = \frac{\zeta^n}{n^s}$$ works.

If you want an example with terms in $$\mathbb{R}$$, just take the real part of the example above: $$a_n = Re(\frac{\zeta^n}{n^s})$$.