How can I be more rigorous in solving the following than just by inspection.
$$\lim_{n\to\infty}\frac{(n+1)^\alpha}{(3n+3)(3n+2)(3n+1)}$$ It's plain to see that if $\alpha>3$ it diverges and for $\alpha\le3$ it converges. I don't know how to show it formally tho. Tried using exp(ln(lim())) to get exponent outside but that didn't get me anywhere. If someone could provide me with idea I would be grateful :)
Hint: $$\begin{align}\lim_{n\to\infty}\frac{(n+1)^\alpha}{(3n+3)(3n+2)(3n+1)}&=\lim_{n\to\infty}\frac{(n+1)^{3}(n+1)^{\alpha-3}}{(3n+3)(3n+2)(3n+1)} \\&=\lim_{n\to\infty}\frac{\frac{1}{n^3}(n+1)^{3}(n+1)^{\alpha-3}}{\frac{1}{n^3}(3n+3)(3n+2)(3n+1)} \\&=\lim_{n\to\infty}\frac{(1+\frac{1}{n})^{3}(n+1)^{\alpha-3}}{(3+\frac{3}{n})(3+\frac{2}{n})(3+\frac{1}{n})} \\&=\frac{1}{27}\lim\limits_{n\rightarrow\infty}(n+1)^{\alpha-3}\end{align}$$