Example 3
Note that this time, you can’t split \((a_n)_n\) up into two monotonic subsequences, so neither of the two methods in the previous examples work. So, we need to be crafty.
It’s always handy to have an idea of what the \(\liminf\) and \(\limsup\) might be. Since \(\left\lvert\cos\left(\frac{n\pi}{3}\right)\right\rvert \leq 1\) for all \(n \in \mathbb{N}\), and \(\frac{(-1)^n}{n} \to 0\) as \(n \to \infty\), we (hopefully) would guess that \[\liminf_{n\to\infty}a_n = -1, \quad \text{and} \quad \limsup_{n \to \infty} a_n = -1.\] So how do we go about showing these?
Solution Firstly, recall that the \(\limsup\) is the largest limit of any subsequence of \((a_n)_n\). Take \(n_j = 6j\), then \[a_{n_j} = \cos\left(\frac{6j\pi}{3}\right) + \frac{(-1)^{6j}}{6j} = 1 + \frac{1}{6j} \to 1, \; \text{as} \; j \to \infty.\]
So, \(\lim_{j \to \infty} a_{n_j} = 1\), hence \(\limsup_{n \to \infty} a_n \geq 1\).
To show that \(\limsup_{n \to \infty} a_n \leq 1\), recall that for sequences \((b_n)_n\) and \((c_n)_n\), \[\limsup_{n \to \infty}(b_n + c_n) \leq \limsup_{n \to \infty}b_n + \limsup_{n \to \infty}c_n.\] Taking \[b_n = \cos\left(\frac{n\pi}{3}\right), \; \text{and} \; c_n = \frac{(-1)^n}{n},\] we have that \[\limsup_{n \to \infty} b_n = 1, \; \text{and}\] \[\limsup_{n \to \infty} c_n = 0 \quad \text{(as $(c_n)_n$ converges to $0$)}.\] Hence, \[\limsup_{n \to \infty} a_n = \limsup_{n \to \infty}(b_n + c_n) \leq 1 + 0 = 1.\]
Combining the two inequalities we have found allows us to conclude that \(\limsup_{n \to \infty} a_n = 1.\)
Have a go at proving that \(\liminf_{n \to \infty} a_n = -1\). You’ll need:
- \(\liminf\) is the smallest limit of any subsequence of \((a_n)_n\).
- \(\liminf_{n \to \infty}(b_n + c_n) \geq \liminf_{n \to \infty}b_n + \liminf_{n \to \infty}c_n.\)