*Notes to a video lecture on http://www.unizor.com*

__Sequence Limit - Infinity__

We use the term

*infinity*rather casually, understanding that this is a large, very large, larger than anything quantity.

This lecture is about a concept of

*infinity*as it is understood by mathematicians.

First of all, let's agree that there is no such number or such quantity as

*infinity*in classical math. There are some advanced parts of higher levels of math, where

*infinity*is introduced as a concrete object, but it is beyond the scope of this course. So, for our purposes

*infinity*is not a number or quantity. What is it then?

**It is a short form of specifying the directional and limitless behavior of a sequence.**

Consider a sequence

**{**that grows boundlessly. That is, for any, however large, number

*X*}_{n}**exists an order number**

*A***such that all members of this sequence with order numbers not less than**

*N***are not less than number**

*N***.**

*A*Using symbols ∀ ("for all" or "for any") and ∃ ("exist"), this can be symbolically written as follows:

∀

**∃**

*A***⇒**

*N: n ≥ N*

*X*_{n}≥ AFor any sequence that behaves in this manner we may say that

**it's limit is infinity**.

Sometimes we add a characteristic "positive" to a word

*infinity*, if it helps to better understand the behavior of a sequence and to differentiate it from

*negative infinity*described below.

So, the expression about

*infinity*(or

*positive infinity*) being a limit of some sequence cannot be considered absolutely rigorous and it just means that the sequence is boundlessly grows in a sense described above. It would be better to use the term

*infinitely growing sequence*than mentioning the word "limit" for such cases. Also not advisable to use the term "convergent" for these sequences, this is reserved for sequences convergent to real numbers.

Example of such infinitely growing sequence:

**{**

*X*}_{n}= 2^{n}Similarly, we can introduce a sequence that boundlessly "grows" (in a sense of absolute value, while being negative) to a

*negative infinity*.

The description of this property is analogous to a case with

*positive infinity*.

If for any negative number

**, however large by absolute value, exists an order number**

*A***such that all members of this sequence with order numbers not less than**

*N***are not greater than number**

*N***, we say that the limit of this sequence is**

*A**negative infinity*.

Symbolically, it can be written as

∀

**∃**

*A***⇒**

*N: n ≥ N*

*X*_{n}≤ A(notice, we don't have to specify that

**is negative since we use "for any" symbol, which includes all negative numbers as well as positive)**

*A*Example of such sequence:

**{**

*X*} = {_{n}= log_{2}(1/n)*−log*}_{2}(n)So, terms

*infinity*,

*positive infinity*(same as

*infinity*) and

*negative infinity*are legitimate mathematical characteristics of sequences that either, being positive starting at some number, grow boundlessly or, while negative after some number, grow by absolute value boundlessly.

Using these terms implies the properties described in detail above. That's why these terms can be considered as a short description of these properties. It's easier and no less rigorous to state "a sequence grows to

*infinity*" instead of "for any, however large, number

**exists an order number**

*A***such that all members of this sequence with order numbers not less than**

*N***are not less than number**

*N***".**

*A*Both expressions mean the same and can be used interchangeably. The former is just a lot shorter and quicker to understand. So is an expression "an infinitely growing sequence".

*Examples*:

1.

*X*_{n}= (n²+1)/nLet's prove that this sequence is limitlessly increasing to

*infinity*.

Choose any boundary number

**, however large. Expression**

*A***is monotonically increasing with an increase of**

*(n²+1)/n=n+1/n***because, as**

*n***increases by**

*n***,**

*1***decreases by a fraction of**

*1/n***. Therefore, once it grows above**

*1***, it will stay above**

*A***. So, we just have to find the first member of this sequence that is above**

*A***.**

*A*Let's find natural order number

**in our sequence for a chosen boundary**

*N***by solving the inequality**

*A*

*(n²+1)/n ≥ A*which is equivalent to (since

*is positive)*

**n**

*n²−An+1 ≥ 0*Expression

*n²−An+1*is a quadratic polynomial of

*with discriminant*

**n****, which is positive for large**

*A²−4**(any value greater than 2).*

**A**This quadratic polynomial limitlessly grows with its argument

**.**

*n*For any large number

**there are two solutions to a quadratic equation**

*A***:**

*n²−An+1 = 0**and*

**n**_{1}= (A−√A²−4)/2

**n**_{2}= (A+√A²−4)/2For any

**greater or equal to a larger of these two solutions**

*N**the inequality we need will be true.*

**n**_{2}Therefore, for any number

**, however large, the members of our sequence**

*A***{**will be greater or equal to this

*(n²+1)/n*}**as long as the order number**

*A***is greater than or equal to**

*n**.*

**(A+√A²−4)/2**This proves that this sequence is limitlessly increasing to

*infinity*.

2.

*X*_{n}= tan(−πn/[2(n+1)])Let's prove that this sequence is limitlessly decreasing to

*negative infinity*.

Choose any boundary number

**, negative and however large by absolute value.**

*A*Let's find

**for a chosen boundary - number**

*N***- by solving the inequality**

*A*

*tan(−πn/[2(n+1)]) ≤ A*which is equivalent to (since

*)*

**tan(−φ) = −tan(φ)**

*−tan(πn/[2(n+1)]) ≤ A*or

*tan(πn/[2(n+1)]) ≥ −A*Here

**is a positive number, however large, function**

*−A***is monotonically increasing on interval**

*tan***[**.

*0,π/2*)Expression

**is monotonically increasing to**

*πn/[2(n+1)]***and its tangent is monotonically increasing to infinity as**

*π/2***is increasing. So, all we have to find is such**

*n***that**

*n***,**

*tan(πn/[2(n+1)]) = −A*which happens at

*πn/[2(n+1)] = arctan(−A)*or

*πn = 2(n+1)arctan(−A)*From the above equation follows:

*n(π−2arctan(−A)) =*

= 2arctan(−A)= 2arctan(−A)

Therefore,

*n = 2arctan(−A) /*

/ (π−2arctan(−A))/ (π−2arctan(−A))

Since

**is a natural number, we have to choose the next natural number greater than the above expression.**

*n*3.

*X*_{n}= n·sin(n)This sequence is growing by absolute value, but changing the sign from positive to negative and back with the periodicity of

*2π*That means, the sequence cannot be qualified as having an

*infinity*(positive or negative) as a limit.

It's not

*infinitely growing*to

*positive infinity*, nor

*infinitely decreasing*to

*negative infinity*.

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