Monotonicity of $a_n=1+\frac{(-1)^n}{n}$

Question

I'm trying to study the monotonicity of $a_n=1+\frac{(-1)^n}{n}$, but what I'm getting isn't correct:

I just assume that $a_n$ is monotonically increasing, and if it isn't, I'll get something absurd:

So, $a_n$ is monotonically increasing iff $$ \begin{split} a_{n+1} \geq a_n &\iff \left(1+\frac{(-1)^{n+1}}{n+1}\right) - \left(1+\frac{(-1)^{n}}{n}\right) \geq 0 \\ &\iff \frac{(-1)^{n+1}}{n+1} - \frac{(-1)^{n}}{n} \geq 0 \\ &\iff \frac{(-1)^{n+1}}{n+1} \geq \frac{(-1)^{n}}{n}, \end{split} $$

Which doesn't make sense since $\lim_{n \to \infty} \frac{(-1)^n}{n}=0$ and $\lim_{n \to \infty} \frac{(-1)^{n+1}}{n+1}=0$, but the later one decreases more rapidly, so $a_n$ would be monotonically decreasing.

But, the solution says that if $n$ is even then $a_n$ is monotonically decreasing and if $n$ is odd then monotonically increasing. How do I prove that? What's wrong with what I did?

Answer 1

HINT There is nothing wrong with your logic, but the last statement is false for any even $n$, since the RHS is positive and the LHS is negative.

To appreciate what is going on, graph the function:

Answer 2

But it's not monotonically increasing because $a_{2n+1}=1-1/(2n+1)<1+1/2n=a_{2n}.$

Answer 3

Your sequence is oscillating around its limit point. If you take a look at two subsequences for odd and even indices, you'll notice that both of them are monotone. One is monotonically increasing, while the other is monotonically decreasing: $$ a_{2k} = 1 + {1\over 2k} \\ a_{2k -1} = 1 - {1\over 2k-1} $$

But the whole sequence is not monotone.

Here is a visualization.

Answer 4

The sequence $(a_n)_{n\geq 1}$ is not monotonic, but the subsequence $(a_{2k})_{k\geq 1}$ is decreasing and the subsequence $(a_{2k+1})_{k\geq 1}$ is increasing.

Answer 5

It is not monotonic since $a_{2n}>1$ and $a_{2n+1}<1$.