# Does sum−1^{n}ln 2n^{frac{1}{n}}) converge or diverge?

Question
Functions
Does $$\displaystyle\sum−{1}^{{{n}}}{\ln{{2}}}{n}^{{{\frac{{{1}}}{{{n}}}}}}{)}$$ converge or diverge?

2021-03-19

Let $$\displaystyle{a}_{{{n}}}={\left(-{1}\right)}^{{{n}}}{\log{{\left({2}{n}^{{{\frac{{{1}}}{{{2}}}}}}\right)}}}.$$ Here
$$\lim_{n \rightarrow \infty}|an| = \lim_{n \rightarrow \infty} \log(2n^{\frac{1}{n}})| =\log(\lim 2n^{\frac{1}{n}}) (log is continuous function in [1,∞) ) =\log2 (Since \lim_{n \rightarrow \infty} n^{\frac{1}{n}}=1$$$$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\left(-{1}\right)}^{{{n}}}{\log{{\left({2}{n}^{{{\frac{{{1}}}{{{n}}}}}}\right)}}}$$ is not convergent.

### Relevant Questions

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