I have the series
$$\sum_{n=2}^\infty \frac{(-i)^n}{\ln n}$$
and I'm trying to determine whether the series will converge or diverge.The Ratio test gives me $L=1$, so it doesn't tell me anything.
I tried to use the comparison test, so I used $\ln n<n$ and thus $\frac{1}{n}<\frac{1}{\ln n}$ for all $n>2$. So I have:
$$|z_n|=\left|\frac{(-i)^n}{\ln n}\right|=\frac{1}{\ln n}\Rightarrow \frac{1}{n}<|z_n|$$ and since $\sum_{n=2}^\infty \frac{1}{n}$ diverges, thus $\sum_{n=2}^\infty |z_n|$ diverges, but I can't say that the series of $z_n$ diverges if the series of absolute values |$z_n|$ diverges. So, how can I show if this series converges or diverges?
Best Answer
Here’s how I would argue:
The real terms and the imaginary terms of the series have nothing to do with each other. The real part is $\sum_{k=1}^\infty\frac{(-1)^k}{\log(2k)}$, while the imaginary part is $\bigl(\sum_{k=0}^\infty\frac{(-1)^k}{\log(2k+3)}\bigr)i$.
More precisely, if $N\equiv1\pmod4$, that is, if $N=4K+1$, then $$ \sum_{n=2}^N\frac{(-i)^n}{\log n}=\sum_{k=1}^{2K}\frac{(-1)^k}{\log(2k)} \>+\>\biggl(\sum_{k=0}^{2K-1}\frac{(-1)^k}{\log(2k+3)}\biggr)i\>, $$ in which I do hope that I have gotten the running indices right. At any rate, the real part and the imaginary part are, separately, alternating series, and there you are.