I have been trying to do this problem by using induction but I became stuck halfway through:
Use induction to show that $$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{n^3}\right) \lt 3 - \frac{1}{n^2}$$ for $n\geq 2$. Does the series $$\sum_{n=1}^{\infty}\frac{1}{n^3}$$ converge? Justify your conclusions.
So far I have this:
Base Case of Induction, $n=2$ $$\begin{align*} \frac{1}{n^3} &\lt 3- \frac{1}{n^2}\\ \frac{1}{8} &\lt 3 - \frac{1}{4}\\ \frac{1}{8} &\lt \frac{11}{4} \end{align*}$$
Induction Step: Assume true for some $k \geq 2 :$ $$ 2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) \lt 3 - \frac{1}{k^2}$$
Show true with $n= k+1$ $$ 2 + 4 + \frac{2}{27} + \cdots + \frac{2}{k^3} + \frac{2}{(k+1)^3} \lt 3-\frac{1}{(k+1)^2}.$$
From there, I have no idea what to do.
I was planning to have $$3 - \frac{1}{k^2} + \frac{2}{(k+1)^3} \lt 3 - \frac{1}{(k+1)^2}$$ but I feel like it won't work since $$2\left(1 + \frac{1}{8} + \frac{1}{27}+\cdots+\frac{1}{k^3}\right)$$ does not equal $$3 - \frac{1}{k^2}.$$