CSIR UGC NET June 2019 Part B Question 22

Question

Answer 1

Let us first note that for $n=2m,m>1$ we have

$$b_n=\frac 12-\sum_{k=3}^n (-1)^k\frac 1k$$

and for $n=2m-1,m>1$ we have

$$b_n=-\sum_{k=3}^n (-1)^k\frac 1k$$

It follows that $\lim b_n\neq 0$. So option1 is not correct.

Now note that 

$$\left(\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2m-1}\right)-\left(\frac{1}{4}+\frac 16+\cdots+\frac{1}{2m}\right)>\left(\frac 13-\frac 14\right)$$

$$\left(\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2m-1}\right)-\left(\frac{1}{4}+\frac 16+\cdots+\frac{1}{2m-2}\right)>0$$

It follows that for $m>1$ we have $$b_{2m-1}>0\quad \text{and}\quad b_{2m}>\frac 12+k,\qquad \text{for some } k>0$$

It follows that $\limsup b_n\geq \frac 12$. Thus Option 2 is a correct choice. 

As $\limsup b_n\geq \frac 12$, the option 4 is also not correct. To argue option 3 is not correct, note that  $\{b_n\}$ is a sequence of nonnegative number and so $\liminf b_n\geq  0$.