Question

Does ${a_n} = {\left( { - \dfrac{1}{2}} \right)^n}$ sequence converge or diverge? How do you find its limit?

Answer 1

Hint: Every infinite sequence is either convergent or divergent. A convergent sequence has a limit that is, it approaches a real number. A divergent sequence doesn’t have a limit.
So, here we will find the limit of the given sequence and then from the limit value we will know whether the given sequence is convergent or divergent.

Complete step-by-step answer:
Given sequence is ${a_n} = {\left( { - \dfrac{1}{2}} \right)^n}$,
Now we will find few terms of the sequence,
First term i.e.,$n = 1$,
$ \Rightarrow {a_1} = {\left( { - \dfrac{1}{2}} \right)^1} = \dfrac{{ - 1}}{2}$,
Second term i.e, $n = 2$,
$ \Rightarrow {a_2} = {\left( { - \dfrac{1}{2}} \right)^2} = \dfrac{1}{4}$,
Third term i.e,$n = 3$,
$ \Rightarrow {a_3} = {\left( { - \dfrac{1}{2}} \right)^3} = \dfrac{{ - 1}}{8}$,
From the terms will can say that the sequence is in a geometric progression with first term $ - \dfrac{1}{2}$ and common ratio i.e.,$r$ is equal to $r = \dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{{\dfrac{1}{4}}}{{\dfrac{{ - 1}}{2}}} = - \dfrac{1}{2}$.
Now the sum of the sequence which is of infinite geometric progression where\[r < 0\]is given by, \[\dfrac{{{a_1}}}{{1 - r}}\],
Now here \[{a_1} = \dfrac{{ - 1}}{2}\] and \[r = \dfrac{{ - 1}}{2}\],
By substituting the values in the formula we get,
\[ \Rightarrow \dfrac{{\dfrac{{ - 1}}{2}}}{{1 - \left( {\dfrac{{ - 1}}{2}} \right)}}\],
Now simplifying we get,
\[ \Rightarrow \dfrac{{\dfrac{{ - 1}}{2}}}{{1 + \dfrac{1}{2}}}\],
Now further simplifying we get,
\[ \Rightarrow \dfrac{{\dfrac{{ - 1}}{2}}}{{\dfrac{3}{2}}}\],
Now eliminating the denominators we get,
Sum of infinite Geometric progression\[ = \dfrac{{ - 1}}{3}\].
Now we have to find whether the sequence converges or diverges,
 Taking the sequence given, i.e., ${a_n} = {\left( { - \dfrac{1}{2}} \right)^n}$,
Now applying limits on both sides we get,
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } {\left( { - \dfrac{1}{2}} \right)^n}$,
Now applying limits by substituting n value we get,
$\mathop { \Rightarrow \lim }\limits_{n \to \infty } {a_n} = - \dfrac{{{{\left( 1 \right)}^\infty }}}{{\left( {{2^\infty }} \right)}}$,
We know that ${1^\infty }$ will be equal to 1 and ${2^\infty }$ will be equal to $\infty $, so the right hand side becomes
$\mathop { \Rightarrow \lim }\limits_{n \to \infty } {a_n} = \dfrac{{ - 1}}{\infty }$,
And we also know that $\dfrac{1}{\infty } = 0$, so finally,
\[\mathop { \Rightarrow \lim }\limits_{n \to \infty } {a_n} = 0\],
From the above we can say that the given sequence converges to 0.

\[\therefore \]The sequence converges to 0 as the limit value of the given sequence i.e., ${a_n} = {\left( { - \dfrac{1}{2}} \right)^n}$ is equal to 0.

Note:
A sequence is like a list of numbers, while a series is a sum of that list. We should remember that a sequence converges if the limit as n approaches infinity of \[{a_n}\] equals a constant number, like 0, 1, pi, or -33. However, if that limit goes to +-infinity, then the sequence is divergent.