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

Question

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.

Answer 1

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 $$ \sum_{k=1}^\infty (-1)^k a_k = \sum_{n=1}^\infty b_n \in\mathbb R.$$

Answer 2

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.

Answer 3

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: $a_n = Re(\frac{\zeta^n}{n^s})$.