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

$\lim_{n\to\infty} \frac{n \sin(n) - n^4}{n^3} \leq \lim_{n\to\infty} \left( \frac{1}{n^2}-n \right)= -\infty\text.$