What is meant by a divergent sequence?
Answer
597k+ views
Hint: Here in this question we have been asked to elaborate the meaning of a divergent sequence. Here in the answer we will define both types of sequence divergent and convergent and then give some examples and elaborate them and give reasons why they are convergent or divergent.
Complete step by step answer:
Now considering from the question we have been asked to elaborate the meaning of a divergent sequence.
Firstly from the basic concepts we know that a sequence ${{a}_{0}},{{a}_{1}},{{a}_{2}},........\in R$ is said to be convergent when there is some $a\in R$ such that ${{a}_{n}}\to a$as $n \to \infty $ and it is said to be divergent when the sequence fails to converge to a finite limit.
For example let us consider a sequence ${{a}_{n}}=n$ which is divergent because ${{a}_{n}}\to \infty $ as $n \to \infty $ .
For example let us consider another sequence ${{a}_{n}}={{\left( -1 \right)}^{n}}$is divergent because it alters between +1 and -1 .
Let us consider another example that is more complex ${{a}_{n}}={{\left( 1+\dfrac{1}{n} \right)}^{n}}$ this sequence converges to $e$ approach can be given as
$\begin{align}
& \displaystyle \lim_{n \to \infty }{{\left( 1+\dfrac{1}{n} \right)}^{n}}=\displaystyle \lim_{n \to \infty }{{e}^{n\ln \left( 1+\dfrac{1}{n} \right)}} \\
& \Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }n\ln \left( 1+\dfrac{1}{n} \right)}}={{e}^{\displaystyle \lim_{n \to \infty }\dfrac{\ln \left( 1+\dfrac{1}{n} \right)}{\dfrac{1}{n}}}} \\
& \Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\dfrac{1}{1+\dfrac{1}{n}}}}=e \\
\end{align}$ .
Any sequence ${{a}_{0}},{{a}_{1}},{{a}_{2}},........\in R$ is said to be convergent with limit $a\in R$ if $\forall \varepsilon >0\exists N\in Z:\forall n\ge N,\left| {{a}_{n}}-a \right|<\varepsilon $ so a sequence can be said that it is divergent if $\forall a\in R\exists \varepsilon >0:\forall N\in Z,\exists n\ge N,\left| {{a}_{n}}-a \right|\ge \varepsilon $ .
Therefore we can conclude that a divergent sequence is a sequence that fails to converge to a finite limit.
Note: This type of questions are completely concept based and can be answered if we have a good understanding of the concept and practice more and more problems related to it. With more practice things get clearer. Possibility of mistakes in this type of questions is very less; it can make mistakes if they are not clear with concepts like interchanging the definitions due to confusion.
Complete step by step answer:
Now considering from the question we have been asked to elaborate the meaning of a divergent sequence.
Firstly from the basic concepts we know that a sequence ${{a}_{0}},{{a}_{1}},{{a}_{2}},........\in R$ is said to be convergent when there is some $a\in R$ such that ${{a}_{n}}\to a$as $n \to \infty $ and it is said to be divergent when the sequence fails to converge to a finite limit.
For example let us consider a sequence ${{a}_{n}}=n$ which is divergent because ${{a}_{n}}\to \infty $ as $n \to \infty $ .
For example let us consider another sequence ${{a}_{n}}={{\left( -1 \right)}^{n}}$is divergent because it alters between +1 and -1 .
Let us consider another example that is more complex ${{a}_{n}}={{\left( 1+\dfrac{1}{n} \right)}^{n}}$ this sequence converges to $e$ approach can be given as
$\begin{align}
& \displaystyle \lim_{n \to \infty }{{\left( 1+\dfrac{1}{n} \right)}^{n}}=\displaystyle \lim_{n \to \infty }{{e}^{n\ln \left( 1+\dfrac{1}{n} \right)}} \\
& \Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }n\ln \left( 1+\dfrac{1}{n} \right)}}={{e}^{\displaystyle \lim_{n \to \infty }\dfrac{\ln \left( 1+\dfrac{1}{n} \right)}{\dfrac{1}{n}}}} \\
& \Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\dfrac{1}{1+\dfrac{1}{n}}}}=e \\
\end{align}$ .
Any sequence ${{a}_{0}},{{a}_{1}},{{a}_{2}},........\in R$ is said to be convergent with limit $a\in R$ if $\forall \varepsilon >0\exists N\in Z:\forall n\ge N,\left| {{a}_{n}}-a \right|<\varepsilon $ so a sequence can be said that it is divergent if $\forall a\in R\exists \varepsilon >0:\forall N\in Z,\exists n\ge N,\left| {{a}_{n}}-a \right|\ge \varepsilon $ .
Therefore we can conclude that a divergent sequence is a sequence that fails to converge to a finite limit.
Note: This type of questions are completely concept based and can be answered if we have a good understanding of the concept and practice more and more problems related to it. With more practice things get clearer. Possibility of mistakes in this type of questions is very less; it can make mistakes if they are not clear with concepts like interchanging the definitions due to confusion.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

