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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Arthur
    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 $l=l'=1$


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