$$\sum_{n=2}^{\infty}{\frac{(-i)^{n}}{\ln n}}$$ In the answer, using comparison test $1/\ln n > 1/n,$ the series is divergent. But, in my opinion, the series can be divided like $$\sum_{n=1}^{\infty}{\frac{(-1)^{n}}{\ln (2n)}}+i\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{\ln (2n+1)}}$$ Then, each series converges, so the series should converge. What is the correct answer?
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2Second one is correct. Your application of Comparison test is wrong. It only shows that the series is not absolutely convergent. – geetha290krm Nov 07 '23 at 07:38
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See also https://math.stackexchange.com/q/3908955/42969 – Martin R Nov 07 '23 at 08:18
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$a_n=\frac 1{\ln n}$ is monotically tends to zero. For $b_n=(-i)^n$ we have $|\sum_{n=2}^N b_n|\leq\sqrt2$ for any $N\geq 2.$ So, according to Dirichlet Test, $$\sum_{n=2}^\infty a_nb_n=\sum_{n=2}^\infty\frac{(-i)^n}{\ln n}$$ converges.
Bob Dobbs
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@geetha290krm His solution is using Dirichlet test two times, mine is using once. – Bob Dobbs Nov 07 '23 at 09:04
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1His proof uses Alternating Series Test (AST) for each of the two terms. The two links provided by Martin R also use AST. – geetha290krm Nov 07 '23 at 09:08