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Limit of a sequence

the sequence limit

Let’s first try to notice some important characteristics of sequences. For sequence:

$ 1, 2, 3, 4,… , n, …$

we can see that members of it are constantly growing, and for sequence

 

$ 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},…,\frac{1}{n},…$

 

we can see that the members are constantly reducing.

Of course, there are many sequences where we can’t find any of that regularities such as alternating sequences.


Sequence of real numbers $ a_1, a_2,…, a_n,…$  will monotonic increase if $ a_n \ge a_{n – 1}$ for every $ n \in \mathbb{N}$.

Sequence of real numbers $ a_1, a_2,…, a_n,…$  will  monotonic decrease if $ a_n \le a_{n – 1}$ for every $ n \in \mathbb{N}$.

The sequence is strictly monotonic if sign $\le$ and $\ge$ is sign $ >$ or $ <$.


Notice that for a proof that a sequence is monotonic increasing it is enough to show that $ a_{n + 1} – a_n \ge 0$

If $ a_{n + 1} – a_n$, then the sequence is decreasing, and if $ a_{n + 1} – a_n$ changes its sign for some values of $n$, then the sequence is not monotonic sequence.

monotonic increasing

monotonic decreasing

Example 1. Is sequence whose general member is $ a_n = \frac{n – 1}{n + 1}$ monotonic?

Let’s find the difference $ a_{n + 1} – a_n  = \frac{n}{n + 2} – \frac{n – 1}{n + 1} = \frac{2}{(n + 2)(n + 1)}$

Since $ n \in \mathbb{N}$, $\frac{2}{(n + 2)(n + 1)} > 0$ or $ a_{n + 1} > a_n$ which means that the sequence is monotonic increasing.

Let’s notice some more properties. For example, what do you notice for a sequence given with general member $ a_n = 1 + \frac{1}{n}$

Probably, just from a look of it, nothing much. Let’s write some of its first members.

$ 1, \frac{3}{2}, \frac{4}{3}, \frac{5}{4}, \frac{6}{5}, \frac{7}{6},…$

We can notice that the members of this sequence never go over $2$, but never decrease below $1$. This means that for every member of this sequence is valid that:

$ 1 < a_n \le 2$

Here we came to second important sequence characteristic:


Sequence $ a_1, a_2,…, a_n,…$ is bounded if there exist numbers $m$ and $M$ such that for every $ a_n$ it is true that:

$ m \le a_n \le M$.

Number $m$ is the lower bound and $M$ is the upper bound of the set $ S = \{ a_1, a_2,…, a_n,…\}$.

We say that the sequence is bounded if we can find its lower and upper bound. Every member of bounded sequence are found in segment $[m, M]$.


If we know that a sequence is bounded, we can find infinitely many upper and lower bounds.

 

Example 2. For the following sequences determin  bound?

a) $ a_n = \frac{1}{n}$

Let’s write again first few members.

$ 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6},…$

monotonic decrease example

We can clearly see that one of the lower bounds is zero and one of the upper is $1$.

$ 0 < a_n \le 1$

b) $ a_n = (- \frac{1}{2})^n$

Don’t let this minus frighten you. Just write first few members and conclude what will happen to this sequence.

$ 1, -\frac{1}{2}, \frac{1}{4}, \frac{1}{16}, -\frac{1}{32}, \frac{1}{64},…$

From here we can see that this sequence is bounded by the first two members. If we’d continue writing these members they would be approaching zero.

This means that $ -\frac{1}{2} < (-\frac{1}{2})^n < \frac{1}{4}$.

c) $ a_n = (- 1)^n n$

Again, we’ll write down first few members.

$ -1, 2, -3, 4, -5, 6, -7…$

writting-down-monotonic-increase-and-decrease-from-example

This sequence has no bounds.

One of the most important questions we can ask when observing infinite sequence $ (a_n)$ is:

What happens to the members of a sequence for great values of $n$?

 

Let’s observe sequence that is given with his general member $ a_n = 1 – \frac{1}{10^n}$.

$ a_1 = 0.9$ $ a_2 = 0.99$ $ a_3 = 0.99$

We can see that the larger the $n$ is, the value of sequences member is closer to $1$.

We say that number 1 is the limit value or limit of a sequence $ a_n$. This statement is noted as:

$ lim_{n -> \infty} (1 – \frac{1}{10^n}) = 1$

To be accurate this is read:

The limit value of sequence $ (1 – \frac{1}{10^n}) = 1$ when $n$ tends to infinity is equal to $1$.

What about alternating sequences? They also can have limits. Let’s observe sequence $ a_n = (- 1)^n \frac{1}{n}$

$ a_1 = – 1$, $ a_2 = \frac{1}{2}$, $ a_3 = – \frac{1}{3}$, $ a_4 = \frac{1}{4}$, $ a_5 = – \frac{1}{5}$, $a_6 = \frac{1}{6}$, …

This sequence is bounded, $ – 1 \le a_n \le 1$ for every n. When n grows members of this sequences are approaching zero and the distance of nth member from zero gets smaller and smaller.

number line sequences greater values

This means that

$ lim_{n -> \infty} (- 1)^n * \frac{1}{n} = 1$


For a sequence $ a_n$ of real numbers is said to be convergent if there exists real number $a$ such that sequence $ a_n$ tends to this number when $n$ grows infinitely.

We say that $a$ is the limit of a sequence and write $ lim_{n -> \infty} (a_n) = a$.


Can we define limit value in some other way?

If we know that that a limit is a number to which the sequence is tending we can rephrase that sentence in some other way. If $a$ is a limit of a sequence, then as $n$ grows, the distance between the member of that sequence and number $a$ is decreasing.

This means that we can find as small number $\varepsilon$ as we’d like, such that the distance from the members of the sequence to number $a$ is smaller than $ε$ for almost every (all but finitely many) member of the sequence or:

$ | a_n – a | < \varepsilon$

Geometrically watched, condition $ | a_n – a | < \varepsilon$ means that there is as small interval as we’d like $ < a – \varepsilon, a + \varepsilon >$ that contains every member but finitely many.


Real number $a$ is the limit of a sequence of real numbers $ a_n$ if for every $\varepsilon > 0$ exists natural number $ n_0$ such that for every $ n > n_0$ it is valid that:

$ | a_n – a | < \varepsilon$


For a sequence that is not convergent we say that it is divergent.

Let’s check that the sequence $ a_n$, $ a_n = \frac{n – 1}{n}$ has limit that is equal to $1$ using this second definition. We can imagine a small number $\varepsilon$, for example $\varepsilon = 0.001$.

$\mid a_n \mid  \le \varepsilon \rightarrow \mid \frac{n – 1}{n} – 1 \mid \le 0.001 \rightarrow \rightarrow = 1000$

From here, we can conclude that for every $ n \ge 1000$ the distance of number $1$ to the member of the sequence will be lesser than $0.001$; every member of a sequence (except for the first 1000) is in environment of number $1$.