Induction Proof: Proof of Strict Inequality involving Exponents of 3 in the Denominator. 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
$$3 - \frac{1}{k^2}.$$

When you go from 
$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) \lt 3 - \frac{1}{k^2}\tag{1}$$
to 
$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}+\frac1{(k+1)^3}\right) \lt 3 - \frac{1}{(k+1)^2}\;,\tag{2}$$
the righthand side increases by
$$\left(3-\frac1{(k+1)^2}\right)-\left(3-\frac1{k^2}\right)=\frac1{k^2}-\frac1{(k+1)^2}\;,$$
while the lefthand side increases by $\dfrac2{(k+1)^3}$. If you can show that $$\frac2{(k+1)^3}\le\frac1{k^2}-\frac1{(k+1)^2}\;,\tag{3}$$ you’ve shown that $(1)$ implies $(2)$, which is your induction step; do you see why?
Now $(3)$ can by multiplied through by $k^2(k+1)^3$ to yield the equivalent inequality $$2k^2\le (k+1)^3-k^2(k+1)\;,$$ which in turn simplifies to $0\le 3k+1$. This is certainly true for $k\ge 2$, and it’s equivalent to $(3)$, so $(3)$ is true for $k\ge 2$, and your induction step is done.
For the second part of the question, note that the $n$-th partial sum of the infinite series is $$\sum_{k=1}^n\frac1{k^3}\;,$$ and from the first part of the question you know that $$2\sum_{k=1}^n\frac1{k^3}<3-\frac1{n^2}\;,$$ so $$\sum_{k=1}^n\frac1{k^3}<\frac32-\frac1{2n^2}\;.$$ Thus, the partial sums form an increasing sequence bounded above by ... what? And what can you conclude from that about the convergence of the infinite series?
