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 $3 - \frac{1}{k^2}.$