Show that the series $$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n -\log n}$$ is conditionally convergent .

My Work

Let $a_n=\frac{(-1)^{n}}{n -\log n}$ then $|a_n|=\frac{1}{n -\log n}$

We can write $$n-\log n \lt n$$ $$\frac{1}{n-\log n} \gt \frac{1}{n}$$

Since $\frac{1}{n}$ diverges ,by comparison test $\sum |a_n|$ also diverges .

But I am not able to prove convergence of $$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n -\log n}$$ .


1 Answer 1


$$\sum_{n=0}^{\infty}\dfrac{(-1)^n}{n-\log n}$$ By Alternating Series test we get$$a_n=\dfrac{1}{n-\ln(n)}$$$$\mbox{$a_n$ is positive and is continuously decreasing from $N=1$}$$$$\lim_{n\rightarrow\infty}\left(\dfrac{1}{n-\ln(n)}\right)=0\implies\mbox{ Converges by Alternating Test}$$


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