Convergence of a sequence
Convergence of sequence can be defined in the two ways: A sequence \(\left\{ {{S_n}} \right\}\) is said to converge to a real number \(l\) , if for each \(\varepsilon > 0\) there exist a positive integer \(m\)(depends upon \(\varepsilon \)) such that
\(\left| {{S_n} - l} \right| < \varepsilon \) for all \(n \ge m\)
Same thing can be explained in the following form:
A sequence \(\left\{ {{S_n}} \right\}\) is said to converge to a real number \(l\)if
\(\mathop {\lim }\limits_{n \to \infty } {S_n} = l\)
Or one may write as \({S_n} \to l\) if \(n \to \infty \)
Let us consider a sequence \(\{ {S_n}\} \), where \({S_n} = \frac{{2n + 1}}{{n + 3}}\)
Here, \(\mathop {\lim }\limits_{n \to \infty } {S_n}\)
\( = \mathop {\lim }\limits_{n \to \infty } \frac{{2n + 1}}{{n + 3}}\)
\( = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{2n + 1}}{n}}}{{\frac{{n + 3}}{n}}}\)
\( = \mathop {\lim }\limits_{n \to \infty } \frac{{2 + \frac{1}{n}}}{{1 + \frac{3}{n}}}\)
\( = \frac{{2 + 0}}{{1 + 0}}\)
\( = 2\)
Therefore, \(\mathop {\lim }\limits_{n \to \infty } {S_n} = 2\) and we can say the sequence converges to the limit 2.
Let us consider another example.
Consider a sequence \(\{ {S_n}\} \), where \({S_n} = \frac{{{n^2}}}{{2n + 3}}\)
Here, \(\mathop {\lim }\limits_{n \to \infty } {S_n}\)
\( = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{2n + 3}}\)
\( = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{2 + \frac{3}{n}}}\)
\( = \infty \)
Here, \(\mathop {\lim }\limits_{n \to \infty } {S_n} = \infty \)
In this case the sequence does not converge to a limit. We say that the sequence diverges to \( + \infty \) .
Convergence of sequence is a vast topic and has tremendous usability in studying the real analysis. If you want to know the basic knowledge about sequence and its related terminologies like bounds of limit, limit point etc., you can follow the link:
Apart from convergence and divergence of sequences one more case arises in the nature of a sequence termed as oscillating sequence.
Oscillating sequence:
A bounded sequence which does not converge and has at least two limit points is said to be an oscillating sequence.For example, consider the sequence \(\{ {S_n} = {( - 1)^n}\} = \left\{ { - 1, + 1, - 1, + 1,...} \right\}\)
Here the sequence \(\{ {S_n}\} \) does not converge to a limit and has two limit points -1 and +1.
Therefore, \(\{ {S_n} = {( - 1)^n}\} = \left\{ { - 1, + 1, - 1, + 1,...} \right\}\) is an oscillating sequence.
Similarly, the sequence \(\{ {S_n} = {( - 1)^n} + 1\} = \left\{ {0,2,0,2,...} \right\}\) is an example of an oscillating sequence. The sequence is not convergent and has two limit points 0 and 2.
Now we will discuss a very useful condition which can be used for determination of convergence of sequences very easily. The condition is given below:
The sequence \(\{ {r^n}\} \) converges if and only if \( - 1 < r \le 1\)
Using the above result we can find some examples of sequences.
1. \(\left\{ {{1^n}} \right\}\) , \(\left\{ {{{\left( {\frac{1}{2}} \right)}^n}} \right\}\) , \(\{ {\left( { - .8} \right)^n}\} \) are few examples of convergent sequence.
2. \(\left\{ {{{1.2}^n}} \right\}\) , \(\left\{ {{{\left( { - \frac{3}{2}} \right)}^n}} \right\}\) , \(\{ {\left( { - 2} \right)^n}\} \) are few examples of non-convergent sequence.
Now we will prove the above statement. To prove the above statement we break down the proof into five cases. For the task, we need to prove the following five cases.
Case 1: Suppose \(r \in (1,\infty )\) or \(r > 1\)
Then r can be written as
\(r = 1 + h\) where \(h > 0\) {e.g. 1.2=1+.2}
\( \Rightarrow {r^n} = {\left( {1 + h} \right)^n} = 1 + nh + \frac{{n(n - 1)}}{{2!}}{h^2} + ...\)
\( \Rightarrow {r^n} = 1 + nh + \frac{{n(n - 1)}}{{2!}}{h^2} + ... \ge 1 + nh\) for all \(n \in N\)
Now, let us consider \(G\) be any number however large,
we have \(1 + nh > G\)
if \(nh > G - 1\)
or \(n > \frac{{G - 1}}{h}\)
Let \(m\) be any positive integer such that \(m > \frac{{G - 1}}{h}\)
Therefore, for \(G > 0\) there exist a positive integer such that
\({r^n} = 1 + nh + \frac{{n(n - 1)}}{{2!}}{h^2} + ... \ge 1 + nh > G\)
\( \Rightarrow {r^n} > G\) for all \(n \ge m\)
Since, the term of the sequences are sufficiently large for \(n \ge m\) , therefore the sequence diverges to \(\infty \) for \(r > 1\).
Case 2: Suppose \(r = 1\)
\( \Rightarrow \left\{ {{r^n}} \right\} = \left\{ {1,1,1,...} \right\}\)
i.e. \(\mathop {\lim }\limits_{n \to \infty } {r^n} = 1\)
Therefore, \(\left\{ {{r^n}} \right\}\) converges for \(r = 1\) .
Case 3: Suppose \(r \in ( - 1,1)\)
Then \(r\) can be written as
\(\left| r \right| = \frac{1}{{1 + h}}\) where \(h > 0\) {e.g. .2=1/(1+4)}
\( \Rightarrow {\left| r \right|^n} = \frac{1}{{{{\left( {1 + h} \right)}^n}}} = \frac{1}{{1 + nh + \frac{{n(n - 1)}}{{2!}}{h^2} + ...}}\)
\( \Rightarrow {\left| r \right|^n} = \frac{1}{{1 + nh + \frac{{n(n - 1)}}{{2!}}{h^2} + ...}} \le \frac{1}{{1 + nh}}\) for all \(n \in N\)
\( \Rightarrow {\left| r \right|^n} \le \frac{1}{{1 + nh}}\) for all \(n \in N\)
Now, let us consider \(\varepsilon \) be any number however large,
we have \(\frac{1}{{1 + nh}} < \varepsilon \)
if \(\frac{{1 - \varepsilon }}{{\varepsilon h}} < n\) ; on simplification.
Let \(m\) be any positive integer such that \(m > \frac{{1 - \varepsilon }}{{\varepsilon h}}\)
Therefore, for \(\varepsilon > 0\) there exist a positive integer \(m\) such that
\( \Rightarrow {\left| r \right|^n} \le \frac{1}{{1 + nh}} < \varepsilon \)
\( \Rightarrow {\left| r \right|^n} < \varepsilon \) for all \(n \ge m\)
\( \Rightarrow {\left| {r - 0} \right|^n} < \varepsilon \) for all \(n \ge m\)
Therefore, the sequence \(\left\{ {{r^n}} \right\}\) converges to the point 0.
Case 4: Suppose \(r = - 1\)
\( \Rightarrow \left\{ {{r^n}} \right\} = \left\{ { - 1, + 1, - 1,...} \right\}\)
i.e. \(\mathop {\lim }\limits_{n \to \infty } {r^n} = - 1\,or\, + 1\)
Therefore, \(\left\{ {{r^n}} \right\}\) has two limit points and it is bounded. Hence it will be an oscillating sequence for \(r = - 1\).
Case 5: Suppose \(r < - 1\)
We consider \(r = - z\) such that \(z > 0\)
\( \Rightarrow \left\{ {{r^n}} \right\} = \left\{ {{{\left( { - z} \right)}^n}} \right\} = \left\{ {{{\left( { - 1} \right)}^n}{{\left( z \right)}^n}} \right\} = \left\{ { - z,{z^2}, - {z^3},...} \right\}\)
Therefore, \(\left\{ {{r^n}} \right\}\) has both positive and negative terms and is also unbounded. Thus, the sequence will be an oscillating sequence.
So, finally we get
From Case 1 \(\left\{ {{r^n}} \right\}\)is divergent for \(r > 1\)
From Case 2 \(\left\{ {{r^n}} \right\}\)is convergent for \(r = 1\)
From Case 3 \(\left\{ {{r^n}} \right\}\)is convergent for \(r \in ( - 1,1)\)
From Case 4 \(\left\{ {{r^n}} \right\}\)is oscillating for \(r = - 1\)
From Case 5 \(\left\{ {{r^n}} \right\}\)is oscillating for \(r < - 1\)
Therefore, finally we can conclude that \(\left\{ {{r^n}} \right\}\) converges if and only if \( - 1 < r \le 1\) .
I hope, this article will help the readers to get a good idea about convergence of a sequence, divergence of a sequence and oscillating sequence. For each definition, examples are given for better understanding about this topic. Finally, an important result has been proved which is used more often for checking of convergence of sequences of the type \(\left\{ {{r^n}} \right\}\).
No comments:
Post a Comment