<s>[INST] 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
$$3 - \frac{1}{k^2}.$$
 [/INST] Hint $\ $ It is simple to prove by induction that an increasing function remains $\ge$ its initial value, i.e. that  $\rm\ f(n+1) \ge f(n)\:$ for $\rm\:n\ge 2\:$ $\:\Rightarrow\:$ $\rm\:f(n) \ge f(2)\:$ for $\rm n\ge 2$.
Now apply that to $\rm\:f(n) = $ RHS $-$ LHS of your inequality, which is increasing since
$$\rm f(n+1) - f(n)\: =\ \frac{3\:n+1}{n^2 (n+1)^3} > 0\ \ \ for\ \ \ n\ge 2\qquad \bf QED$$
Remark $\ $ It's worth mention that this may be viewed as prototypical example of telescopy. Indeed, viewing $\rm\:f(n)\:$ as the sum of its first differences $\rm\:f(n) = f(2) + \sum_{k=2}^{n-1} (f(k+1)-f(k)),\:$ the proof reduces to a trivial induction that a sum is $> 0\:$ if each summand is. In a similar way one may exploit telescopy to simplify many inductive proofs to trivial inductions (e.g. the trivial induction that $\rm 1^n = 1$). For many examples of this technique see my prior posts on telescopy.
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