Can convergent series have a non-exist or infinity entry?

Question

In my advanced mathematical analysis class I came across this question: determine whether the below series converges. $$\sum\frac{1}{n \sqrt{\log n}}$$ As you can see when $n=1$, the sequence $a_n$ does not exist. Or should we say it's infinity? So my question is, when one of the entries in a series is infinity, can the series still be convergent? The given solution is that this series is divergent, by the integral test. But I still don't understand how, if you have $$\sum\frac{1}{n \sqrt{\log n}} = \infty + \frac{1}{2 \sqrt{\log2}} + ...$$ the result can be finite.

Answer 1

The series is reasonably intended to start at some $n_0$ such that all the terms are actually defined.
Of course $$ \sum_{n\geq 2}\frac{1}{n\sqrt{\log n}} $$ is divergent, either by applying Cauchy's condensation test or by noticing that

$$ 2\sqrt{\log(n+1)}-2\sqrt{\log n}=2\frac{\log\left(1+\frac{1}{n}\right)}{\sqrt{\log(n+1)}+\sqrt{\log n}}\leq \frac{1}{n\sqrt{\log n}} $$ such that $$ \sum_{n=2}^{N}\frac{1}{n\sqrt{\log n}}\geq 2\sqrt{\log(N+1)}-2\sqrt{\log 2} . $$