How can I prove that $\sum_n \left(\frac{n+1}{n^2+1}\right)^3$ converges?

Question

Clearly $\sum_n \left(\frac{n+1}{n^2+1}\right)^3$ converges, but I am having an embarrassingly hard time proving it.

I tried the Cauchy Root test, leading me to prove that $\limsup \left(\frac{n+1}{n^2+1} \right)^\frac{3}{n}=1$, implying that the root test cannot be used. Since the ratio test is equivalent to the root test, I cannot use that either.

I also had the idea of showing that $\frac{2}{n^3}>\left(\frac{n+1}{n^2+1}\right)^3$ for all $n \in \mathbb{N}$, but I have no idea how to show that.

Any suggestions?

Answer 1

I solved it. I had to first prove $0< \left(\frac{n+1}{n^2+1}\right)^3 < \frac{8}{n^3}$. By induction, I showed $-n^2+n < 2$ for all $n \in \mathbb{N}$. You can algebraically manipulate $-n^2+n < 2$ to get $\left(\frac{n+1}{n^2+1}\right)^3 < \frac{8}{n^3}$. It is also clear that $0<\left(\frac{n+1}{n^2+1}\right)^3< \frac{8}{n^3}$. Since $\sum_n \frac{1}{n^3}$ converges, then $\sum_n \frac{8}{n^3}$ converges. Therefore since $0<\left(\frac{n+1}{n^2+1}\right)^3 < \frac{8}{n^3}$ and $\sum_n \frac{8}{n^3}$ converges, then by the Comparison test $\sum_n \left(\frac{n+1}{n^2+1}\right)^3$ converges.

Answer 2

Another useful technique in those cases is the limit comparison test with $\sum_{n=1}^{\infty}\dfrac{1}{n^{\alpha}}$ for suitable $\alpha$. Sometimes limits are more immediate than inequalities.