<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] It doesn't matter that the sum does not equal $ 3- \frac{1}{k^3}$. What matters is that the sum is less than $3 - \frac{1}{k^3}$.
We have:
$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) \lt 3 - \frac{1}{k^2}.$$
Adding $\frac{2}{(k+1)^3}$ to both sides does not change the inequality, so we have
$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) + \frac{2}{(k+1)^3} \lt 3 - \frac{1}{k^2} + \frac{2}{(k+1)^3}.$$
The left hand side is the quantity you want. 
With the right hand side, we have
$$\frac{1}{k^2}\gt \frac{1}{(k+1)^2},$$
so
$$-\frac{1}{k^2} \lt -\frac{1}{(k+1)^2},$$
hence
$$\frac{2}{(k+1)^3} - \frac{1}{k^2} \lt \frac{2}{(k+1)^3} - \frac{1}{(k+1)^2} = \frac{1}{(k+1)^2},$$
so
$$3 - \frac{1}{k^2}+\frac{2}{(k+1)^3} \lt 3 - \frac{1}{(k+1)^2}.$$
And since $a\lt b$ and $b\lt c$ implies $a\lt c$, then...
For the last part of the question, note that the quantity on the left hand side is twice the $n$th partial sum of 
$$\sum_{n=2}^{\infty} \frac{1}{n^3}.$$
So you should be able to conclude that the partial sums of 
$$\sum_{n=1}^{\infty}\frac{1}{n^3}$$
are bounded above; since all terms are positive, that means that...
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