# Ideas to know if $\sum_{2\leq n\leq x}\frac{(-1)^n\zeta(n)}{\operatorname{rad}(n)}$ has finite limit as $x\to\infty$

Let for integers $n\geq 1$ the radical of an integer, see the definition of this arithmetical function, for example, from this Wikipedia.

I wondered next question when I did a comparison with a well-known identity from the literature.

Question. Does converge $$\sum_{n=2}^\infty\frac{(-1)^n\zeta(n)}{\operatorname{rad}(n)},\tag{1}$$ where $\zeta(s)$ denotes the Riemann's Zeta function? Thanks in advance.

Then we need to know relevant facts about the asymptotic behaviour of $$\sum_{2\leq n\leq x}\frac{(-1)^n\zeta(n)}{\operatorname{rad}(n)},\tag{2}$$ as $x\to\infty$. I believe that there are two approaches, using Abel's summation identity or well with summation by parts. I tried the first one.

From Abel's summation formula one knows that $$\sum_{2<n\leq y}\frac{(-1)^n\zeta(n)}{\operatorname{rad}(n)}=\left(\sum_{1\leq n\leq y}\frac{(-1)^n}{\operatorname{rad}(n)}\right)\zeta(y)-\frac{\zeta(2)}{2}-\int_2^y\left(\sum_{1\leq n\leq t}\frac{(-1)^n}{\operatorname{rad}(n)}\right)\zeta'(t)dt.\tag{3}$$

Also we know that $$\lim_{y\to\infty}\zeta(y)=1.\tag{4}$$

To solve previous question I require an easy way while it is possible (do not use the advanced knowledges). If you follow my way, we need to state facts about the asymptotic behaviour of $$\sum_{1\leq n\leq y}\frac{(-1)^n}{\operatorname{rad}(n)},$$ but as I've said to me isn't required the best statement. Just those ideas and calculations to know if our series in $(1)$ will be convergent. Of course you can to choose yourself strategy.

No, this does not converge: the terms do not even converge to $0$. For any $m\geq 1$, the term for $n=2^m$ is $\frac{\zeta(2^m)}{2}> \frac{1}{2}$, so there are infinitely many terms greater than $\frac{1}{2}$.