# Operations and limits with sequences

## Operations with sequences

Real numbers allow us to define the operations of sum, subtraction, multiplication and division. These operations can be naturally extended to the set of the sequences. The way of extending the operations is performed term-wise. Let's see the corresponding definitions.

Let $$a=\{a_n\}_{n\in\mathbb{N}}$$ and $$b=\{b_n\}_{n\in\mathbb{N}}$$ be two sequences. We define the following sequences: $$a+b=\{a_n+b_n\}_{n\in\mathbb{N}}$$$$$a-b=\{a_n-b_n\}_{n\in\mathbb{N}}$$$ $$a\cdot b=\{a_n\cdot b_n\}_{n\in\mathbb{N}}$$$Also, we can define the sequence $$\dfrac{a}{b}$$ like $$\Big\{\dfrac{a_n}{b_n}\Big\}_{n\in\mathbb{N}}$$, but in this case it is necessary to demand $$b_n\neq0$$ so that the division is definite. Let's see some example so we have no doubt about it. Let's say $$a=\{n\}_{n\in\mathbb{N}}$$ and $$b=\{\dfrac{1}{n} \}{n\in\mathbb{N}}$$. Then, $$a+b=\Big\{n+\dfrac{1}{n} \Big\}_{n\in\mathbb{N}}=\Big\{\dfrac{n^2+1}{n}\Big\}_{n\in\mathbb{N}}$$$ $$a-b=\Big\{n-\dfrac{1}{n} \Big\}_{n\in\mathbb{N}}=\Big\{\dfrac{n^2-1}{n}\Big\}_{n\in\mathbb{N}}$$$$$a\cdot b=\Big\{n\cdot\dfrac{1}{n} \Big\}_{n\in\mathbb{N}}=\{1\}_{n\in\mathbb{N}}$$$ $$a/b=\Big\{\dfrac{n}{1/n} \Big\}_{n\in\mathbb{N}}=\{n^2\}_{n\in\mathbb{N}}$$$## Limits and operations with convergent sequences It is natural now to wonder how the sequences limit behave with respect to the operations. In this sense, the limit acts as simply as possible when the sequences are convergent. $$\lim_{n \to \infty}{(a\pm b)}_n=\lim_{n \to \infty}{a_n}\pm\lim_{n \to \infty}{b_n}$$$ $$\lim_{n \to \infty}{(a\cdot b)}_n=\lim_{n \to \infty}{a_n}\cdot\lim_{n \to \infty}{b_n}$$$$$\lim_{n \to \infty}{(a/b)}_n=\lim_{n \to \infty}{a_n}/\lim_{n \to \infty}{b_n}$$$

The latter property needs that $$\lim_{n \to \infty}{b_n}\neq0$$.

These properties allow us to calculate the limit through already well-known limits. The following proposition turns out to be still more useful: The limit of the product of a bounded sequence by another with limit zero has limit zero.

Let's see an example of this proposition.

We consider the sequence $$a_n=(-1)^n$$, remember that this sequence does not have an extreme but it is bounded, and the sequence $$b_n=\dfrac{1}{n}$$, which has limit $$0$$. According to the previous proposition the product of the two sequences has limit $$0$$. That is to say: $$\lim_{n \to \infty}{\dfrac{(-1)^n}{n}}=0$$$## Limits and operations in general If we accept some arithmetical rules with the infinite then we can extend the previous rules to divergent sequences. Let $$a$$ be a real number, we define the following operations: $$a\pm\infty=\pm\infty$$$ $$\infty+\infty=\infty$$$$$-\infty-\infty=-\infty$$$ $$\dfrac{a}{+\infty}=0$$$$$\infty\cdot(-\infty)=(-\infty)\cdot\infty=-\infty$$$ $$\infty\cdot\infty=(-\infty)\cdot(-\infty)=\infty$$$If $$a > 0$$ we also define $$a\cdot(\pm\infty)=(\pm\infty)\cdot a=\pm\infty$$ And if $$a < 0$$ then • $$a\cdot\infty=\infty\cdot a=-\infty$$ • $$a\cdot(-\infty)=(-\infty)\cdot a=\infty$$ These arithmetic rules extend the operations with sequences limits, for both convergent and divergent sequences. Let's see some examples. Let's say $$a=\{n\}_{n\in\mathbb{N}}$$ and $$b=\Big\{\dfrac{2n+1}{n}\Big\}{n\in\mathbb{N}}$$. We verify the previous rules for the operation of these sequences. For adding and subtraction $$\lim_{n \to \infty}{\Big(n\pm\dfrac{2n+1}{n}\Big)}=\lim_{n \to \infty}{\Big(\dfrac{n^2\pm2n\pm1}{n}\Big)}=\infty$$$ $$\lim_{n \to \infty}{n}\pm\lim_{n \to \infty}{\dfrac{2n+1}{n}}=\infty\pm2=\infty$$$For the product $$\lim_{n \to \infty}{\Big(n\cdot\dfrac{2n+1}{n}\Big)}=\lim_{n \to \infty}{(2n+1)}=\infty$$$ $$\lim_{n \to \infty}{n}\cdot\lim_{n \to \infty}{\dfrac{2n+1}{n}}=\infty\cdot2=\infty$$$For the division $$\dfrac{a}{b}$$ $$\lim_{n \to \infty}{\Big(n/\dfrac{2n+1}{n}\Big)}=\lim_{n \to \infty}{\Big(\dfrac{n^2}{2n+1}\Big)}=\infty$$$ $$\lim_{n \to \infty}{n}/\lim_{n \to \infty}{\dfrac{2n+1}{n}}=\dfrac{\infty}{2}=\infty$$$For the division $$\dfrac{b}{a}$$ $$\lim_{n \to \infty}{\Big(\dfrac{2n+1}{n}/n\Big)}=\lim_{n \to \infty}{\Big(\dfrac{2n+1}{n^2}\Big)}=0$$$ $$\lim_{n \to \infty}{\dfrac{2n+1}{n}}/\lim_{n \to \infty}{n}=\dfrac{2}{\infty}=0$$$## Indeterminate forms The previous arithmetical rules allow us to define most of the operations with sequence limits. If we notice, all the possible operations are described except for $$\infty-\infty$$, $$\dfrac{\pm\infty}{\infty}$$ and $$0\cdot(\pm\infty)$$. Common sense makes us think that the results should be $$0$$, $$\pm1$$ and $$0$$ respectively. We are going to see that this is not true in general. We consider the sequences with the general term $$n+1$$ and $$n$$. Both have infinite limit but let's see the limit of the difference. $$\lim_{n \to \infty}{(n+1-n)}=\lim_{n \to \infty}{1}=1$$$

Considering the sequences with general term $$2n$$ and $$n$$ we see how the limit of the quotient is not $$1$$.

$$\lim_{n \to \infty}{\dfrac{2n}{n}}=\lim_{n \to \infty}{2}=2$$$For the last case we choose the sequences with general term $$n$$ and $$\dfrac{1}{n}$$. We see that the limit of the product is $$1$$. $$\lim_{n \to \infty}{n\cdot\dfrac{1}{n}}=\lim_{n \to \infty}{1}=1$$$

These operations can give any number or infinity.

To solve this indecisiveness of the indeterminate form there is no generic method and every case must be studied separately. For the cases where the sequences are a polynomial quotient, it is sufficient to manipulate the expressions until finding an expression as quotient of two polynomials and applying the already well-known theory. Let's see a more complete example than the previous ones.

We consider the sequences $$a=\Big\{\dfrac{n^2}{n+1}\Big\}_{n\in\mathbb{N}}$$ and $$b=\Big\{\dfrac{2n^2}{n-1}\Big\}_{n\in\mathbb{N}}$$, both with infinite limit.

We calculate the limit of $$a-b$$:

$$\lim_{n \to \infty}{\Big(\dfrac{n^2}{n+1}-\dfrac{2n^2}{n-1}/n\Big)}=\lim_{n \to \infty}{\dfrac{n^2(n-1)-2n^2(n+1)}{(n+1)(n-1)}}=\lim_{n \to \infty}{\dfrac{-n^3-3n^2}{n^2-1}}=-\infty$$$In the first equality we have put a common denominator and in the second one we have simplified the expression. We calculate the limit of $$\dfrac{a}{b}$$: $$\lim_{n \to \infty}{\Big(\dfrac{n^2}{n+1}/\dfrac{2n^2}{n-1}/n\Big)}=\lim_{n \to \infty}{\dfrac{n^2(n-1)}{2n^2(n+1)}}=\lim_{n \to \infty}{\dfrac{n^3-n^2}{2n^3+2n^2}}=\dfrac{1}{2}$$$

The first equality corresponds to performing the crossed product and the second one to simplifying the expression.

Let's also see the following example with the sequences $$a=\Big\{\dfrac{n^2-3}{n}\Big\}_{n\in\mathbb{N}}$$ and $$b=\Big\{\dfrac{5n^2}{7n^3-n}\Big\}_{n\in\mathbb{N}}$$ with limits infinte and zero, respectively.

We calculate the limit of the product $$a\cdot b$$:

$$\lim_{n \to \infty}{\Big(\dfrac{n^2-3}{n}\cdot\dfrac{5n^2}{7n^3-n}\Big)}=\lim_{n \to \infty}{\dfrac{5n^4-15n^2}{7n^4-n^2}}=\dfrac{5}{7}$$\$