Question

Determine whether the sequence converges or diverges. If it converges, find the limit.

\({a_n} = \frac{{{{\cos }^2}n}}{{{2^n}}}\)

Short answer

Converges &\(\mathop {\lim }\limits_{n \to \infty } {a_n} = 0\)

Step-by-step solution

Definition

A sequence\(\left\{ {{a_n}} \right\}\)has the limit \(L\)and we write \(\mathop {\lim }\limits_{n \to \infty } {a_n} = L\;\;or\;\;{a_n} \to L\)as\(n \to \infty \)if we can make the terms\({a_n}\)as close to\(L\)as we like by taking\(n\)sufficiently large. If \(\mathop {\lim }\limits_{n \to \infty } {a_n}\)exists, we say the sequence converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent).

Simplify

Consider the sequence\({a_n} = \frac{{{{\cos }^2}n}}{{{2^n}}}\).

We know that\( - 1 \le \cos x \le 1\)

Then \(0 \le {\cos ^2}x \le 1\)for all values of\(x\).

Since \({a_n} = \frac{{{{\cos }^2}n}}{{{2^n}}}\)then \(\frac{0}{{{2^n}}} \le \frac{{{{\cos }^2}n}}{{{2^n}}} \le \frac{1}{{{2^n}}}\)for all values of\(n\).

Apply squeeze theorem

Consider \(\frac{0}{{{2^n}}} \le \frac{{{{\cos }^2}n}}{{{2^n}}} \le \frac{1}{{{2^n}}} = 0 \le \frac{{{{\cos }^2}n}}{{{2^n}}} \le \frac{1}{{{2^n}}}\) for all values of\(n\).

The Squeeze theorem states that:

If \({a_n} \le {b_n} \le {c_n}\)for \(n \ge {n_0}\)and\(\mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } {c_n} = L\)then \(\mathop {\lim }\limits_{n \to \infty } {b_n} = L\)

Since \(0 \le \frac{{{{\cos }^2}n}}{{{2^n}}} \le \frac{1}{{{2^n}}}\) for all values of\(n\).

Consider \(\mathop {\lim }\limits_{n \to \infty } 0 = 0\) and\(\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{2^n}}} = 0\)then by Squeeze theorem\(\mathop {\lim }\limits_{n \to \infty } \frac{{{{\cos }^2}n}}{{{2^n}}} = 0\).

Hence the sequence \({a_n} = \frac{{{{\cos }^2}n}}{{{2^n}}}\) is convergent and\(\mathop {\lim }\limits_{n \to \infty } {a_n} = 0\).