Question

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

Answer 1

The infinite series $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + .....$ is an elementary example of a geometric series that converges absolutely. The sum of the series is $1$ . In summation notation, this may be expressed as $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + ..... = \sum\limits_{n = 1}^\infty {{{\left[ {\dfrac{1}{2}} \right]}^n}} = 1$.
After finding the geometric progression of the sequence , we can find out the values needed to find the limits of the sequence. Here we will find the sequence converges or diverges by finding the common ratio and then put the limit hence find the result.

Complete step by step solution:
The given equation is
${a_n} = {\left[ { - \dfrac{1}{2}} \right]^n}$
Let us consider the few terms of this sequence
When $n = 1$
$\Rightarrow {a_1} = \left[ { - \dfrac{1}{2}} \right]$
When $n = 2$
$\Rightarrow {a_2} = \left[ {\dfrac{1}{4}} \right]$
When $n = 3$
$\Rightarrow {a_3} = \left[ { - \dfrac{1}{8}} \right]$
This is the geometric progression of first term ${a_1} = \left[ { - \dfrac{1}{2}} \right]$
And the common ratio $(r) = \left[ { - \dfrac{1}{2}} \right]$
Consider the limit where $n \to \infty$
$\mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } {\left[ { - \dfrac{1}{2}} \right]^n} = 0$
$\therefore$ The sequence will converge.
The sum of the infinite series where $r < 0$ is given by
$\sum\limits_{n = 1}^\infty {{a_n}} = \dfrac{{{a_1}}}{{1 - r}}$
Now substituting the values of ${a_1}$ and $r$ , we get
$\Rightarrow \sum\limits_{n = 1}^\infty {{a_n}} = \dfrac{{\left[ { - \dfrac{1}{2}} \right]}}{{1 - \left[ { - \dfrac{1}{2}} \right]}}$
Now solving the denominator,
$\Rightarrow \sum\limits_{n = 1}^\infty {{a_n}} = \dfrac{{\left[ { - \dfrac{1}{2}} \right]}}{{\left[ {\dfrac{3}{2}} \right]}}$
Solving the above equation by bring $\left( {\dfrac{3}{2}} \right)$ upside, we get
$\Rightarrow \sum\limits_{n = 1}^\infty {{a_n}} = \left[ { - \dfrac{1}{3}} \right]$
$\therefore$ The given sequence converges and the sum of the infinite sequence converges to $- \dfrac{1}{3}$

Note:
We say that a sequence converges if the sequence has a finite limit $L$ ,the sequence then has convergence, it converges to the limit $L$ and we describe the sequence as convergent. If a sequence is convergent, then its limit is unique. On the other hand, if the limit of a sequence grows without bound in either the positive or negative direction the sequence is said to diverge. The sequence has divergence and we describe the sequence as divergent. Keep in mind that being divergent is not the same as not
having a limit. The limit of a sequence ${a_n}$ is the number $L$ such that if for each $\varepsilon > 0$, there exists an
integer $N$ such that $|{a_n} - L| < \varepsilon$ for all $n > N$. $|{a_n} - L| < \varepsilon$ means the values of ${a_n}$ such
that $L - \varepsilon < {a_n} < L + \varepsilon$.
Each sequence’s limit falls under only one of the four possible cases:
A limit exists and the limit is $L:\,\mathop {\lim }\limits_{n \to + \infty } {a_n} = L$
There is no limit:$\mathop {\lim }\limits_{n \to + \infty } {a_n}$does not exist.
The limit grows without bound in the positive direction: $\mathop {\lim }\limits_{n \to + \infty } {a_n} = + \infty$
The limit grows without bound in the negative direction: $\mathop {\lim }\limits_{n \to + \infty } {a_n} = - \infty$ .