limit of real number sequence

An endless real number sequence

$a_1,a_2,a_3,\mathrm{\dots }$(1)

has the real number $A$ as its limit, if the distancebetween $A$ and $a_n$ can be made smaller than an arbitrarilysmall positive number $\epsilon$ by chosing the $n$ of $a_n$ sufficiently great, i.e. greater than a number $N$ (the of which depends on the value of $\epsilon$); accordingly

Then we may denote

$\underset{n\to \mathrm{\infty }}{lim}{a}_n=A$(2)

or equivalently

$a_n\to A\mathit{\hspace{1em}}\text{as}\mathit{\hspace{1em}}n\to \mathrm{\infty }.$(3)

Remark 1.  One should not think, that  $a_n=A$  when  $n=\mathrm{\infty }$.  The symbol “$\mathrm{\infty }$” no number, one cannot set it for the value of $n$.  It’s only a question of allowing $n$ to exceed any necessaryvalue.

Example 1.  Using the notation (2) we can write a result

$$\underset{n\to \mathrm{\infty }}{lim}\frac{2n}{n+1}=\mathrm{\hspace{0.33em}2}.$$

It’s a question of that the real number sequence

$$\frac{2}{2},\frac{4}{3},\frac{6}{4},\mathrm{\dots }$$

has the limit value 2 (e.g. the nine hundred ninety-ninth member  $\frac{1998}{1000}=1.998$  is already “almost” 2!).  For justificating the result, let $\epsilon$ be an arbitrary positive number, as small as you want.  Then

(4)

when $n$ is chosen so big that

$n>{\displaystyle \frac{2}{\epsilon }}-1.$(5)

The condition (5) is obtained from (4) by solving this inequality for $n$.  In this case, we have $N=\frac{2}{\epsilon }-1$.

Example 2.  The so-called decimal expansions, i.e. endless decimal numbers, such as

$3.14159265\mathrm{\dots }=\pi ,0.636363\mathrm{\dots },0.99999\mathrm{\dots },$(6)

are, as a matter of fact, limits of certain real number sequences.  E.g. the last of these is related to the sequence

$0.9,\mathrm{\hspace{0.33em}0.99},\mathrm{\hspace{0.33em}0.999},\mathrm{\dots }$(7)

which may be also written as

$$1-\frac{1}{10},\mathrm{\hspace{0.33em}1}-\frac{1}{{10}^2},\mathrm{\hspace{0.33em}1}-\frac{1}{{10}^3},\mathrm{\dots }$$

The limit of (7) is 1.  Actually, if  $\epsilon >0$,  the distance between 1 and the $n^{\mathrm{th}}$ member of (7) is

when  ${10}^n>\frac{1}{\epsilon }$,  i.e. when  $n>-{\log }_{10}\epsilon =N$.

The endless decimal notations (6) and others are, in fact, limit notations — no finite amount of decimals in them suffices to give their exact values.

Remark 2.  In both of the above examples, no of the sequence members was equal to the limit, but it does not need always to be so; thus for example

$$\underset{n\to \mathrm{\infty }}{lim}\frac{1+{(-1)}^n}{2n}=0$$

and every other member of the sequence in question is 0.

Infinite limits of real number sequences

There are sequences that have no limit at all, for example  $1,-1,\mathrm{\hspace{0.17em}1},-1,\mathrm{\hspace{0.17em}1},-1,\mathrm{\dots }$.  Some real number sequences (1) have the property, that the member $a_n$ may exeed every beforehand given real number $M$ if one takes$n$ greater than some value $N$ (which depends on $M$):

$$a_n>M\mathit{\hspace{1em}}\text{when}\mathit{\hspace{1em}}n>N.$$

Then we write

$$\underset{n\to \mathrm{\infty }}{lim}{a}_n=\mathrm{\infty }.$$

Similarly, the sequence (1) may be such that for each positive $M$ there is $N$ such that

and then we write

$$\underset{n\to \mathrm{\infty }}{lim}{a}_n=-\mathrm{\infty }.$$

E.g.

$$\underset{n\to \mathrm{\infty }}{lim}{n}^2=\mathrm{\infty },\underset{n\to \mathrm{\infty }}{lim}(1-n)=-\mathrm{\infty }.$$