Question

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

\({a_n} = \frac{{{{( - 1)}^n}}}{{2\sqrt n }}\)

Short answer

The sequence converges and the value of limit is 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

Given a sequence \({a_n} = \frac{{{{( - 1)}^n}}}{{2\sqrt n }}\).

Evaluate\(\mathop {\lim }\limits_{n \to \infty } |{a_n}|\) as:

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{( - 1)}^n}}}{{2\sqrt n }}} \right| &= \frac{1}{2}\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{\sqrt n }}} \right)\\ &= \frac{1}{2} \cdot 0\\ &= 0\end{aligned}\)

Compute\(\mathop {\lim }\limits_{n \to \infty } {a_n}\)

Now using the theorem:

If\(\mathop {\lim }\limits_{n \to \infty } \left| {{a_n}} \right| = 0\), then\(\mathop {\lim }\limits_{n \to \infty } {a_n} = 0\)

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } {a_n} &= \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{( - 1)}^n}}}{{2\sqrt n }}} \right)\\ &= 0\end{aligned}\).

So, the sequence \({a_n} = \frac{{{{( - 1)}^n}}}{{2\sqrt n }}\) converges and the value of limit is 0.