Question

Show that the series whose $n$th term is $\frac{\sqrt[3]{32n^2-1}}{\sqrt[4]{43n^3+2n+5}}$ is divergent.

Answer 1

To show that the series whose nth term is

$\frac{\sqrt[3]{32n^2-1}}{\sqrt[4]{43n^3+2n+5}}$ is divergent

Here,

$a_n=\frac{\sqrt[3]{32n^2-1}}{\sqrt[4]{43n^3+2n+5}}$

As $n$ increases, $a_n$ approximates to the value

$a_n=\frac{\sqrt[3]{32n^2}}{\sqrt[4]{43n^3}}$

This can also be written as

$a_n=\frac{\sqrt[3]{32}}{\sqrt[4]{43}}\times \frac{1}{n^{\frac{1}{12}}}$

If $b_n=\frac{1}{n^{\frac{1}{12}}}$

We have,

$\underset{n\to \infty }{lim}\frac{a_n}{b_n}=\frac{\sqrt[3]{32}}{\sqrt[4]{43}}$ which is finite

Therefore, the series whose nth term is $\frac{1}{n^{12}}$ may be taken as the auxiliary series.

But this series is Divergent.

Hence, given series is divergent.