Question

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

\({a_n} = {2^{ - n}}\cos n\pi \)

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).

Evaluate limit

Consider the sequence:

\(\begin{aligned}{a_n} &= {2^{ - n}}\cos n\pi \\ &= \frac{{\cos n\pi }}{{{2^n}}}\end{aligned}\)

Hence, we can find the limit:

\(\left\{{\begin{aligned}{\mathop {\lim }\limits_{n \to \infty }(\cos n\pi ) \in ( - 1,1)}\\{\mathop {\lim }\limits_{n \to \infty} \left( {{2^n}} \right) = \infty }\end{aligned}} \right.\)

And then\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\cos n\pi }}{{{2^n}}}} \right) = 0\)

So the sequence converges to zero as\(n \to \infty \).