Question

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

\({a_n} = \frac{{{n^2}}}{{\sqrt {{n^3} + 4n} }}\)

Short answer

The 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).

Simplify

Given a sequence\({a_n} = \frac{{{n^2}}}{{\sqrt {{n^3} + 4n} }}\).

\(\begin{aligned}{a_n} &= \frac{{{n^2}}}{{\sqrt {{n^3} + 4n} }}\\ &= \frac{{\frac{{{n^2}}}{{\sqrt {{n^3}} }}}}{{\frac{{\sqrt {{n^3} + 4n} }}{{\sqrt {{n^3}} }}}}\,\,\,\,\,\,\,\,({\rm{Dividing}}\,{\rm{Numerator}}\,\& \,{\rm{denominator}}\,{\rm{by}}\,\sqrt {{n^3}} )\\ &= \frac{{\sqrt n }}{{\sqrt {1 + \frac{4}{{{n^2}}}} }}\end{aligned}\)

Evaluate limit

Now evaluate the limit as\(n \to \infty \).

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt n }}{{\sqrt {1 + \frac{4}{{{n^2}}}} }} &= \frac{\infty }{1}\\ &= \infty \end{aligned}\)

Which is infinite.

Hence the sequence \({a_n} = \frac{{{n^2}}}{{\sqrt {{n^3} + 4n} }}\) diverges.