Question

What is the sum \[1 + \dfrac{1}{3} + \dfrac{1}{5} + \dfrac{1}{7} + \dfrac{1}{9} + \dfrac{1}{{11}} + ...\] ?

Answer 1

Hint: In the above given problem, we are given a series of positive terms as \[1 + \dfrac{1}{3} + \dfrac{1}{5} + \dfrac{1}{7} + \dfrac{1}{9} + \dfrac{1}{{11}} + ...\] . Since the given series has an infinite number of terms, hence it is an infinite series. We have to find the sum of the above given infinite series. In order to approach the required solution, first we have to determine if the given series is convergent or non-convergent i.e. divergent. The sum of the series exists as a finite value if and only if the nature of the series is converging. Otherwise, if we have the series as a divergent series then the sum of the series does not exist i.e. it is undefined or infinity.

Complete answer:
Given series is, \[1 + \dfrac{1}{3} + \dfrac{1}{5} + \dfrac{1}{7} + \dfrac{1}{9} + \dfrac{1}{{11}} + ...\]
Let the given infinite series be \[S\] i.e.
\[ \Rightarrow S = 1 + \dfrac{1}{3} + \dfrac{1}{5} + \dfrac{1}{7} + \dfrac{1}{9} + \dfrac{1}{{11}} + ...\]
We have to find the sum of the series \[S\] .
If we notice each term of the series \[S\] , we can write the nth term of the series as,
\[ \Rightarrow {T_n} = \dfrac{1}{{2n - 1}}\]
Therefore, we can also write the series in form of,
\[ \Rightarrow S = \sum\limits_{n = 1}^\infty {\dfrac{1}{{2n - 1}}} \]
Now applying the limit comparison test of the above series with respect to the harmonic series \[\sum\limits_{n = 1}^\infty {\dfrac{1}{n}} \] , we can write
\[ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{\dfrac{1}{{2n - 1}}}}{{\dfrac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{n}{{2n - 1}}\]
Taking \[n\] common from the denominator, we get
\[ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{n}{{2n - 1}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{n}{{n\left( {2 - \dfrac{1}{n}} \right)}}\]
That gives us,
\[ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{n}{{n\left( {2 - \dfrac{1}{n}} \right)}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{2 - \dfrac{1}{n}}}\]
Applying the limit \[\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n} = 0\] now, we can write the above equation as,
\[ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{2 - \dfrac{1}{n}}} = \dfrac{1}{2}\]
Since the obtained limit is finite and non-zero, \[\dfrac{1}{2} \ne 0\] , and the harmonic series \[\sum\limits_{n = 1}^\infty {\dfrac{1}{n}} \] is a divergent series therefore we can conclude that the give series \[ \Rightarrow S = \sum\limits_{n = 1}^\infty {\dfrac{1}{{2n - 1}}} \] is also a divergent series.
Therefore, the series \[1 + \dfrac{1}{3} + \dfrac{1}{5} + \dfrac{1}{7} + \dfrac{1}{9} + \dfrac{1}{{11}} + ...\] is a divergent series and the sum of this series is not defined.

Note: The limit comparison test is used to determine the convergence of an infinite series \[\sum {{a_n}} \] by comparing the limits of the series \[\sum {{a_n}} \] and \[\sum {{b_n}} \] where \[\sum {{b_n}} \] is already known to be convergent or divergent.
According to the limit comparison test, if the limit \[\mathop {\lim }\limits_{n \to \infty } \dfrac{{{a_n}}}{{{b_n}}} \ne 0\] and is finite, then both the series \[\sum {{a_n}} \] and \[\sum {{b_n}} \] have the same nature of convergence or divergence.