Let \(a_{n}= (-1)^{n}\log(2n^{\frac{1}{2}}).\) Here

\(\lim_{n \rightarrow \infty}|an| = \lim_{n \rightarrow \infty} \log(2n^{\frac{1}{n}})| =\log(\lim 2n^{\frac{1}{n}}) (log is continuous function in [1,∞) ) =\log2 (Since \lim_{n \rightarrow \infty} n^{\frac{1}{n}}=1\)

\(\sum_{n=1}^\infty (-1)^{n}\log(2n^{\frac{1}{n}})\) is not convergent.

\(\lim_{n \rightarrow \infty}|an| = \lim_{n \rightarrow \infty} \log(2n^{\frac{1}{n}})| =\log(\lim 2n^{\frac{1}{n}}) (log is continuous function in [1,∞) ) =\log2 (Since \lim_{n \rightarrow \infty} n^{\frac{1}{n}}=1\)

\(\sum_{n=1}^\infty (-1)^{n}\log(2n^{\frac{1}{n}})\) is not convergent.