Question

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

\({a_n} = \frac{{{{\left( { - 1} \right)}^n}n}}{{n + \sqrt n }}\)

Short answer

Sequence diverges.

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

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

Apply the limit test to the nth term of the sequence is as follows:

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } {a_n} &= \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^{n + 1}}n}}{{n + \sqrt n }}\\ &= \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {1/n} \right)}}{{\left( {1/n} \right)}}\frac{{{{\left( { - 1} \right)}^{n + 1}}n}}{{n + \sqrt n }}{\rm{, Since }}\frac{{\left( {1/n} \right)}}{{\left( {1/n} \right)}} &= 1{\rm{ for }}n \ne 0\\ &= \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{1 + \frac{1}{{\sqrt n }}}}\\ &= \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}{\rm{, Since }}\frac{1}{{\sqrt n }} \to 0\end{aligned}\)

But \({\left( { - 1} \right)^{n + 1}}\)oscillates between\( \pm 1\), never converges.

Thus sequence \({a_n} = \frac{{{{\left( { - 1} \right)}^n}n}}{{n + \sqrt n }}\)diverges.