Question

Does \[{a_n} = \dfrac{1}{{{n^2} + 1}}\] converge?

Answer 1

Hint: In the above question, we are given a sequence \[{a_n} = \dfrac{1}{{{n^2} + 1}}\] where \[n\] is any natural number. We have to find if the given sequence is converging or diverging. We must use the definition of convergence of a sequence which states that:
If for all \[\varepsilon > 0\] there exists a natural number \[m\]
Such that, for all \[n \geqslant m\] implies that \[\left| {{a_n} - a} \right| < \varepsilon \]
Then \[{a_n}\]converges to \[a\] i.e. \[\mathop {\lim }\limits_{n \to \infty } {a_n} = a\] .
Mathematically,
If \[\forall \varepsilon > 0\] \[\exists m \in \mathbb{N}\]
Such that, \[\forall n \geqslant m \Rightarrow \left| {{a_n} - a} \right| < \varepsilon \]
Then \[\mathop {\lim }\limits_{n \to \infty } {a_n} = a\]

Complete step-by-step answer:
Given sequence is \[{a_n} = \dfrac{1}{{{n^2} + 1}}\]
We have to find the convergence of \[{a_n}\] .
Let \[\dfrac{1}{{{n^2} + 1}} < \varepsilon \] for some constant \[\varepsilon > 0\]
Then, taking reciprocal of both sides, we get
\[ \Rightarrow {n^2} + 1 > \dfrac{1}{\varepsilon }\]
Subtracting \[1\] from both sides,
\[ \Rightarrow {n^2} > \dfrac{1}{\varepsilon } - 1\]
Taking square roots of both sides,
\[ \Rightarrow n > \sqrt {\dfrac{1}{\varepsilon } - 1} \]
Therefore, \[\dfrac{1}{{{n^2} + 1}} < \varepsilon \] when \[n > \sqrt {\dfrac{1}{\varepsilon } - 1} \] .
Since \[\sqrt {\dfrac{1}{\varepsilon } - 1} \] is a constant, let a constant natural number \[m\] such that \[m > \sqrt {\dfrac{1}{\varepsilon } - 1} \] .
Now we can write it in the form of the definition as follows:
For all \[\varepsilon > 0\] there exists a natural number \[m\]
Such that \[\forall n \geqslant m\] \[ \Rightarrow \dfrac{1}{{{n^2} + 1}} < \varepsilon \]
So, subtracting zero from L.H.S.
\[ \Rightarrow \dfrac{1}{{{n^2} + 1}} - 0 < \varepsilon \]
Hence, taking modulus of L.H.S. we get
\[ \Rightarrow \left| {\dfrac{1}{{{n^2} + 1}} - 0} \right| < \varepsilon \]
Thus, the definition of convergence is satisfied.
Hence, \[\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{{n^2} + 1}} = 0\] .
Therefore, by the definition of convergence of sequence we can say that the sequence \[{a_n}\] is convergent and it converges to \[0\] , i.e. \[\mathop {\lim }\limits_{n \to \infty } {a_n} = 0\] .

Note: A sequence is called convergent if it converges to a finite value in the range of real numbers for any however large natural number. However, if the value of the sequence keeps increasing or decreasing for however large natural number towards an infinite range, \[ + \infty \] or \[ - \infty \] , then the sequence is said to be non-convergent or divergent and it diverges to \[ + \infty \] or \[ - \infty \] .
The definition of a divergent sequence can be given as:
For any however large real number \[h\] there exists a natural number \[m\]
Such that, for all natural numbers \[n \geqslant m\] it implies that \[{a_n} > h\] .
Then, \[\mathop {\lim }\limits_{n \to \infty } {a_n} = \infty \]
And if it implies that \[{a_n} < - h\]
Then, \[\mathop {\lim }\limits_{n \to \infty } {a_n} = - \infty \]