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Question: What are two examples of convergent sequences?...

What are two examples of convergent sequences?

Explanation

Solution

We solve this problem by using the definition of convergence and some tests used to find the convergence. There are mainly 2 tests we use for writing the examples of convergent sequences.
(1) p – series test:
If the sequence in the form 1np\sum{\dfrac{1}{{{n}^{p}}}} then,
If p<1p<1 then the sequence converges and if p1p\ge 1 then the sequence diverges.
(2) Root test:
If the nth{{n}^{th}} term of a sequence is an{{a}_{n}} and l=limnannl=\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{\left| {{a}_{n}} \right|}
If l<1l < 1 then the sequence converges and if l1l\ge 1 then the sequence diverges.

Complete step by step answer:
We are asked to give two examples of convergent sequences.
Let us use two tests for convergence to find the examples of convergent sequences.
First let us use the p – series test.
We know that the p – series test is given as,
If the sequence in the form 1np\sum{\dfrac{1}{{{n}^{p}}}} then,
If p<1p < 1 then the sequence converges and if p1p\ge 1 then the sequence diverges.
Now, let us take an example where p<1p < 1 in the sequence 1np\sum{\dfrac{1}{{{n}^{p}}}} then we get one example as 1n\sum{\dfrac{1}{\sqrt{n}}}
Here, we can see that in 1n\sum{\dfrac{1}{\sqrt{n}}} the value of p'p' is 12\dfrac{1}{2} which is less than 1.
Now, let us use the root test.
We know that the root test is given as,
If the nth{{n}^{th}} term of a sequence is an{{a}_{n}} and l=limnannl=\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{\left| {{a}_{n}} \right|}
If l<1l < 1 then the sequence converges and if l1l\ge 1 then the sequence diverges.
Let us assume that the nth{{n}^{th}} term of sequence as,
an=13n\Rightarrow {{a}_{n}}=\dfrac{1}{{{3}^{n}}}
Now, by using the root test to above sequence then we get,
l=limn13nn l=limn13=13 \begin{aligned} & \Rightarrow l=\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{\dfrac{1}{{{3}^{n}}}} \\\ & \Rightarrow l=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{3}=\dfrac{1}{3} \\\ \end{aligned}
Here, we can see that the value of l'l' is less than 1 so that the sequence is convergent.
Therefore, we can conclude that the two examples of convergent sequence are given as,
(1) 1n\sum{\dfrac{1}{\sqrt{n}}}
(2) 13n\sum{\dfrac{1}{{{3}^{n}}}}

Note: The main mistake is done in p – series test. The p – series test have the summation as 1np\sum{\dfrac{1}{{{n}^{p}}}} then we can compare the value of p'p' with respect to ‘1’ to get the nature of sequence.
But some students may take the p – series as np\sum{{{n}^{p}}} and compare the value of p'p' which gives the wrong answer.