Question

What will be the limit of the following sequence as \[n\] tends to infinity? Will the sequence converge or diverge?
\[\mathop {\lim }\limits_{n \to \infty } {\text{ }}{a_n}\] where \[{a_n} = {\left( {1 + \sin n} \right)^{\dfrac{1}{n}}}\]

Answer 1

Hint:In this problem, we have to find the limit of the function and check whether the function converges or diverges. In order to find the limit, we will substitute \[\infty \] at the place of \[n\] and using the concepts \[ - 1 \leqslant \sin x \leqslant 1\] and \[{a^0} = 1\] we will get the required limit. After that we will check whether the limit converges or diverges by using the concept that the sequence \[\left\{ {{a_n}} \right\}\] converges if \[\mathop {\lim }\limits_{n \to \infty } {\text{ }}{a_n}\] exists and having a finite limit otherwise it diverges. And hence, we will get the required result.

Complete step by step answer:
We have the given sequence as
\[{a_n} = {\left( {1 + \sin n} \right)^{\dfrac{1}{n}}}\]
Now taking limit on both sides, we get
\[\mathop {\lim }\limits_{n \to \infty } {\text{ }}{a_n} = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \sin n} \right)^{\dfrac{1}{n}}}\]
On substituting \[\infty \] at the place of \[n\] we have
\[\mathop {\lim }\limits_{n \to \infty } {\text{ }}{a_n} = {\left( {1 + \sin \infty } \right)^{\dfrac{1}{\infty }}}\]
Now we know that
\[\dfrac{1}{\infty } = 0\]
Therefore, we have
\[\mathop {\lim }\limits_{n \to \infty } {\text{ }}{a_n} = {\left( {1 + \sin \infty } \right)^0}\]
Now we know \[ - 1 \leqslant \sin x \leqslant 1\]
It means that the value of \[\sin \infty \] is any number between \[ - 1\] and \[1\]
Therefore, we can write
\[\mathop {\lim }\limits_{n \to \infty } {\text{ }}{a_n} = {\left( {1 + \left( {any{\text{ }}number{\text{ }}between{\text{ }} - 1{\text{ }}and{\text{ }}1} \right)} \right)^0}\]
Now we know that
Any number raised to the power \[0\] is equal to \[1\] i.e., \[{a^0} = 1\]
Therefore, we have
\[\mathop {\lim }\limits_{n \to \infty } {\text{ }}{a_n} = 1\]
Hence, the limit of the given sequence \[{a_n} = {\left( {1 + \sin n} \right)^{\dfrac{1}{n}}}\] equals to \[1\]
Now we know that the sequence \[\left\{ {{a_n}} \right\}\] converges if \[\mathop {\lim }\limits_{n \to \infty } {\text{ }}{a_n}\] exists and has a finite limit otherwise it diverges.
As we can see that the above \[\mathop {\lim }\limits_{n \to \infty } {\text{ }}{a_n}\] exists and has a finite value.
Therefore, \[\mathop {\lim }\limits_{n \to \infty } {\text{ }}{a_n}\] converges
Hence, the given sequence is convergent, and it converges to \[1\].

Note:Students make mistakes while substituting the limit values and finding the final answer. Also students get confused while finding the value of \[\sin \infty \] .So, always remember that the range of \[\sin x\] is \[\left[ { - 1,1} \right]\] .Also one point to note while solving the problems related to limit is that if you get any indeterminate form, then first convert it into the normal form and then simplify further