# Show that the sequence $x_n=\left[1+\frac{(-1)^n}{n}\right]$ is convergent using monotone convergence theorem.

Question: Show that the sequence $x_n=\left[1+\frac{(-1)^n}{n}\right]$ is convergent using monotone convergence theorem.

My Attempt: I can show that $x_1=0$,$x_2=\frac{3}{2}$,$x_3=\frac{2}{3}$,$x_4=\frac{5}{4}$,... and hence conclude that $x_n \le \frac{3}{2}$ intuitively. But can anybody state a rigorous method? What about monotonicity? I can see it is not monotonic but oscillating. But it still converges. How?

• Look at $x_{2n} = 1 + \frac{1}{2n}$ and $x_{2n+1} = 1-\frac{1}{2n+1}$. These are both monotonic, and both are bounded (in the relevant direction) by $1$, so they must both converge. I don't think that you can use the monotonic convergence theorem to prove that these two subsequences converge to the same number, though. There you would have to use some other argument. Nov 11 '15 at 14:17

One has $x_{2n+1}\leq 1\leq x_{2n}$ and $u_n=x_{2n+1}$ is increasing while $v_n=x_{2n}$ is decreasing.
Indeed $u_{n+1}-u_n={1\over 2n+1}-{1\over 2n+3}\geq 0$ and $v_{n+1}-v_n={1\over 2n+2}-{1\over 2n}\leq 0$
Monotone convergence tells us that $u_n\to l\leq 1$ and $v_n\to l'\geq 1$
$$|u_n-v_n|={1\over 2n+1}+{1\over 2n}={4n+1\over 4n^2+2n}\to 0$$
And so $l=l'=1$