Question

Find whether the series in which
$u_n=\sqrt[3]{3n^3+1}-n$
is convergent or divergent.

Answer 1

Here $u_n=n(\sqrt[3]{31+\frac{1}{n^3}}-1)$
$=n(1+\frac{1}{3n^3}-\frac{1}{9n^6}+.....-1)$
$=\frac{1}{3n^2}-\frac{1}{9n^5}+.....$
If we take $v_n=\frac{1}{n^2}$, we have
$\frac{u_n}{v_n}=\frac{1}{3}-\frac{1}{9n^5}+..$
$\therefore Lim=\frac{u_n}{v_n}=\frac{1}{3}$
But the auxiliary series
$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...\frac{1}{n^2}$
is convergent, therefore the given series is convergent.