Check convergence $\sum\limits_{n=1}^\infty\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$

Question

Please help me to check convergence of $$\sum_{n=1}^{\infty}\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$$

Answer 1

Hint: Consider the Taylor expansion of the summand about $1/n$ for $n$ large:

$$\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}} \approx \left (1 + \frac{7}{2 n^2} \right ) - \left (1 - \frac{8}{3 n^2} \right ) = \frac{37}{16 n^2}$$

Use the comparison test.

Answer 2

(Presumably with tools from Calc I...) Using the conjugate identities $$ \sqrt{a}-1=\frac{a-1}{1+\sqrt{a}},\qquad1-\sqrt[3]{b}=\frac{1-b}{1+\sqrt[3]{b}+\sqrt[3]{b^2}}, $$ one gets $$ \sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}=x_n+y_n, $$ with $$ x_n=\sqrt{1+a_n}-1=\frac{a_n}{1+\sqrt{1+a_n}},\qquad a_n=\frac{7}{n^2}, $$ and $$ y_n=1-\sqrt[3]{1-b_n}=\frac{b_n}{1+\sqrt[3]{1-b_n}+(\sqrt[3]{1-b_n})^2},\qquad b_n=\frac{8}{n^{2}}-\frac{1}{n^{3}}. $$ The rest is easy: when $n\to\infty$, the denominator of $x_n$ is at least $2$ hence $x_n\leqslant\frac12a_n$, likewise the denominator of $y_n$ is at least $1$ hence $y_n\leqslant b_n$. Thus, $$ 0\leqslant x_n+y_n\leqslant\tfrac12a_n+b_n\leqslant\frac{12}{n^2}, $$ since $a_n=\frac7{n^2}$ and $b_n\leqslant\frac8{n^2}$. The series $\sum\limits_n\frac1{n^2}$ converges, hence all this proves that the series $\sum\limits_nx_n$ converges (absolutely).

Answer 3

Update. Compare $$\sum_{n=1}^{\infty}\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right) $$ with the serie $$ \sum_{n=1}^{\infty} \frac{1}{n^a} $$ for a convenient $a\geq 1$. Note that if $a_n=\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$, $b_n=\frac{1}{n^a}$ and
$$ 0<\lim_{n\to\infty} \frac{a_n}{b_n}<\infty. $$ then $\sum_{n=1}^{\infty} a_n<\infty$ if, only if, $\sum_{n=1}^{\infty} b_n<\infty$.