Converging series and converging alternative series implies absolute convergence?

Question

It is known that that harmonic series diverges, but the alternating form of the harmonic series converges. However, I am not sure if there are examples of series $a_n$ that

$$ \sum_{n=1}^{\infty} a_n \space \text{converges} \space \text{and} \space \sum_{n=1}^{\infty} (-1)^{n+1}a_n \space \text{also converges}$$ but the series does not converge absolutely. A candidate $a_n$ I can think of is $$ \sum_{n=1}^{\infty} \frac{(-1)^n}{n}$$, but this may be problematic because the signs do not agree. Is there more inspiring examples?

Answer 1

That candidate doesn't work, because$$\sum_{n=1}^\infty(-1)^{n+1}\frac{(-1)^n}n=\sum_{n-1}^\infty-\frac1n,$$which diverges.

However, that property holds for the series\begin{multline}1+1+(-1)+(-1)+\frac12+\frac12+\left(-\frac12\right)+\left(-\frac12\right)+\\+\frac13+\frac13+\left(-\frac13\right)+\left(-\frac13\right)+\cdots\end{multline}

Answer 2

You can modify your example by taking complex values: Let $x= \exp(2 \pi it)$ for some $t \in (0,\pi)$ and consider the series representation of the main branch of the logarithm, which by Abel-summation converges also on the boundary of the unit ball - except in zero. We have $$-\log(1-x) = \sum_{k=1}^\infty \frac{x^k}{k}$$ and $$\log(1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1} x^k}{k}.$$ Both series are convergent, but we don't have absolutely convergence.