Estimation

Limits part 1: Infinite sequences: Techniques

Estimation

The first few values of the sequence \(a_n = (-1)^n + n\) are \[ 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, \dots \] For even values of \(n\) we have \(a_n = n - 1\) and for odd values of \(n\) it holds that \(a_n = n+1\). Have a look at this animation.

We see that \(a_n \geq n - 1\) for all \(n\). Since the sequence is greater than a sequence that diverges to \(\infty\) we conclude that \(a_n\) has to diverge to \(\infty\) as well. The following theorem states exactly this:

Let \((a_n)_{n=1}^\infty\) and \((b_n)_{n=1}^\infty\) be two sequences such that \(a_n \geq b_n\) for all \(n\). Then we have \[ \lim_{n\to\infty} a_n \geq \lim_{n\to\infty} b_n \] if these limits exist.

If \(\lim_{n\to\infty} b_n = \infty\) then it is not necessary to assume existence of \(\lim_{n\to\infty} a_n\) because it always exists. and the value of this limit has to be \(\infty\).

There is a similar result for \(-\infty\):

Let \((a_n)_{n=1}^\infty\) and \((b_n)_{n=1}^\infty\) be two sequences such that \(a_n \leq b_n\) for all \(n\). Then we have \[ \lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n \] if these limits exist.

These two propositions are especially useful if you suspect that the limit of a sequence equals \(\pm\infty\).

We have \[ \lim_{n\to\infty} \frac{n \sin(n) - n^4}{n^3} = -\infty\text.\]

Note that \(-1\leq \sin(n) \leq 1\). Hence for each \(n\) it is true that \[ \frac{n \sin(n) - n^4}{n^3} \leq \frac{n \cdot 1 - n^4}{n^3} = \frac{n-n^4}{n^3} = \frac{1}{n^2}-n\text.\] The calculation rules for limits lead to \[ \lim_{n\to\infty} \left( \frac{1}{n^2}-n \right)= 0 -\infty = -\infty\text.\] The above theorem now implies that \[ \lim_{n\to\infty} \frac{n \sin(n) - n^4}{n^3} \leq \lim_{n\to\infty} \left( \frac{1}{n^2}-n \right)= -\infty\text. \]