Product of converging series

Question

$\displaystyle \sum_{n=1}^∞ (-1)^n\dfrac{1}{n}.\dfrac{1}{2^n}$

Knowing that

1. An alternating harmonic series is always convergent
2. Riemann series are always convergent when $p>1$

Is it safe to say that the product of these two is convergent (as described above)?

Answer 1

hint: Your series is absolutely convergent because $|a_n| \leq \left(\dfrac{1}{2}\right)^n$. And if you have $2$ converging series, say $\displaystyle \sum_{k=1}^\infty a_k, \displaystyle \sum_{k=1}^\infty b_k$, then their product as you define it yourself $\displaystyle \sum_{k=1}^\infty a_kb_k$ may or may not converge depending on the general terms $a_k,b_k$. You can take for example $a_k = b_k = (-1)\dfrac{1}{\sqrt{k}}$, and you see each is conditionally convergent hence each is convergent, but their product by your way is a diverging harmonic series.

Answer 2

@Kf-Sansoo @Joanpemo

In fact, in my opinion, the right "product" here is the dot product $U.V$ in Hilbert space $\ell^2$, which is meaningful because

$$U=(-1, \dfrac{1}{2},-\dfrac{1}{3},\cdots(-1)^n \dfrac{1}{n},\cdots) \ \ and \ \ V=(1,\dfrac{1}{2},\dfrac{1}{4},\cdots\dfrac{1}{2^n}\cdots)$$

are square sommable with respective squared norm $\dfrac{\pi^2}{6}$ (classical sum) and $\dfrac{4}{3}$ (sum of a geometric series with ratio $\dfrac{1}{4}$).

Besides, there is another reason for which this series a sommable: it is the value for $x=1/2$ of the entire series $\sum_{n=1}^{\infty}\dfrac{(-x)^n}{n}$ which is known to be $ln(1+x)$; therefore, the value of the dot product is $ln(1+\dfrac{1}{2})=ln(\dfrac{3}{2})$.