convergence of $\sum_{n=1}^{\infty}(\sqrt{n^2+7}-\sqrt[3]{n^3+8n+1})\ln(1+1/n)$

Question

$$\sum_{n=1}^{\infty}(\sqrt{n^2+7}-\sqrt[3]{n^3+8n+1})\ln(1+1/n)$$ I eventually reached $\sum(n(\sqrt{1+7/n^2}-\sqrt[3]{1+8/n^2+1/n^3})\ln(1+1/n))$ and I think this is a dead end. I have no other ideas how to deal with it.

Edit: I just thought that Dirichlet test is the key, but I'm not sure how to use it.

Answer 1

I eventually reached $(\ast)=\sum\limits_n n\left(\sqrt{1+7/n^2}-\sqrt[3]{1+8/n^2+1/n^3}\right)\ln(1+1/n)$ and I think this is a dead end.

Not at all! To continue without limited expansions, since this is a requisite of the question, note the following:

• $(1+4/n^2)^2\geqslant1+8/n^2\geqslant1+7/n^2$ hence $1\leqslant\sqrt{1+7/n^2}\leqslant1+4/n^2$
• $(1+3/n^2)^3\geqslant1+9/n^2\geqslant1+8/n^2+1/n^3$ hence $1\leqslant\sqrt[3]{1+8/n^2+1/n^3}\leqslant1+3/n^2$
• $0\leqslant\ln(1+1/n)\leqslant1/n$

Thus, $$0\leqslant n\left(\sqrt{1+7/n^2}-1\right)\ln(1+1/n)\leqslant4/n^2,$$ and $$0\leqslant n\left(\sqrt[3]{1+8/n^2+1/n^3}-1\right)\ln(1+1/n)\leqslant3/n^2,$$ from which the (absolute) convergence of the series $(\ast)$ follows.

Answer 2

$u_n=\{(n^2+7)^\frac{1}{2}-(n^3+8n-7)^\frac{1}{3}\}\ln \frac{n+1}{n}$

$=\{n(1+\frac{7}{n^2})^\frac{1}{2}-n(1+\frac{8}{n^2}+\frac{1}{n^3})^\frac{1}{3}\}\ln\frac{n+1}{n}$

$=\{n(1+\frac{1}{2}\frac{7}{n^2}+....)-n(1+\frac{1}{3}(\frac{8}{n^2}+\frac{1}{n^3})+...\}\ln\frac{n+1}{n}$

Comparing $u_n $ with $v_n=\frac{1}{n^2}$ we get $\lim_{n\rightarrow \infty}\frac{u_n}{v_n}=\frac{7}{2}-\frac{8}{3}=\frac{5}{6}\neq 0$

Hence the series converges

also use the fact that $\lim \frac{\ln(1+\frac{1}{n})}{\frac{1}{n}}=1$