# Show that the given series is conditionally convergent.

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 $$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n -\log n}$$ .

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