Proof that $\frac{9}{8} < \sum\limits_{n=1}^{\infty} \frac{1}{n^3} < \frac{5}{4}$. I'm trying to prove that $\frac{9}{8} < \sum\limits_{n=1}^{\infty} \frac{1}{n^3} < \frac{5}{4}$. I've seen similar proofs to this that tend to approach the proofs geometrically, using the upper and lower bounds of the remainder, the sum of the series less its $k$th partial sum, generated from the integral test. It usually involves some trick like "excluding $k$ terms." My professor tends to start at the second term to find a "lower bound," from which the proof follows with algebraic manipulation, but seemed to suggest to me that this followed from trial and error. 
So, with that said, I really can't say that I understand the intuition behind finding or proving this. The first step seems to be plotting $y = \frac{1}{x^3}$ and then considering upper and lower Riemann sums (geometrically, "boxes" above the curve that will generate a sum greater than the area below it and boxes below the curve that will generate a sum less than the area below it). Though I believe I can draw the graph, I'm struggling with how to approach the problem, even if I have the abstract idea right, which I  quite doubt.
I'd very much appreciate if someone could shed some light on this. Thanks in advance. 
 A: Approach Avoiding Integrals
Note that
$$
\begin{align}
\frac1{\left(n-\frac12\right)^2}-\frac1{\left(n+\frac12\right)^2}
&=\frac{2n}{\left(n^2-\frac14\right)^2}\\
&\gt\frac2{n^3}
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{n=1}^\infty\frac1{n^3}
&\lt1+\frac12\sum_{n=2}^\infty\left[\frac1{\left(n-\frac12\right)^2}-\frac1{\left(n+\frac12\right)^2}\right]\\
&=\frac{11}9
\end{align}
$$
The other direction is simply
$$
\begin{align}
\sum_{n=1}^\infty\frac1{n^3}
&\gt\sum_{n=1}^2\frac1{n^3}\\
&=\frac98
\end{align}
$$
Thus, we get the tighter bounds
$$
\frac98\lt\sum_{n=1}^\infty\frac1{n^3}\lt\frac{11}9
$$

Bounding by Integrals
Note  that
$$
\int_n^{n+1}\frac1{x^3}\,\mathrm{d}x\le\frac1{n^3}\le\int_{n-1}^n\frac1{x^3}\,\mathrm{d}x
$$
Therefore,
$$
1+\overbrace{\int_2^\infty\frac1{x^3}\,\mathrm{d}x}^{\le\sum\limits_{k=2}^\infty\frac1{n^3}}\le\sum_{n=1}^\infty\frac1{n^3}\le1+\frac18+\overbrace{\int_2^\infty\frac1{x^3}\,\mathrm{d}x}^{\ge\sum\limits_{k=3}^\infty\frac1{n^3}}
$$
Thus,
$$
\frac98\le\sum_{n=1}^\infty\frac1{n^3}\le\frac54
$$
A: 
See green bars for LH inequality and orange bars for RH inequality. 
$$\begin{align}
1+\int_2^\infty\frac 1{x^3}\; dx 
\quad &<\quad 
\sum_{n=1}^\infty \frac 1{n^3} 
 &&<\quad
1+\frac 18+\int_2^\infty \frac 1{x^3} \; dx\\
\frac 98 
\quad &<\quad
\sum_{n=1}^\infty \frac 1{n^3} 
 &&<\quad\frac 54
&
\end{align}$$
A: $$\sum_{n=1}^{\infty} \frac{1}{n^3} =\sum_{n=1}^k \frac{1}{n^3} + \sum_{n=k+1}^{\infty} \frac{1}{n^3}< \sum_{n=1}^k \frac{1}{n^3}+\int_{k}^{\infty} \frac{1}{x^3} dx =\sum_{n=1}^k \frac{1}{n^3}+\frac{1}{2 k^2}$$
and this for any $k\ge 1$. For $k=2$ we get RHS $=1+1/8+1/8=5/4$.
A: Your intuition to attack this problem with a geometrical approach and recognizing the usefulness of the integral test in similar problems is good. Once you suspect that the integral test is going to be useful, a great way to confirm this is to draw out a few boxes of height $\frac{1}{n^3}$ starting at $1$ and ask yourself if there are easy to integrate monotonically decreasing functions that bound these boxes below or above.
Once you do this, it's clear that $\int_{k}^\infty \frac{1}{x^3} dx$ bounds the sum $\sum_{n = k}^\infty \frac{1}{n^3}$ below and the sum $\sum_{n= k + 1}^\infty \frac{1}{n^3}$ above. The rest of the problem is working out how many terms in the sum you need to take to make this bound tight enough to satisfy the prompt.
A: Here's another approach:
\begin{align}\frac{9}{8} =\frac{1}{1^3} + \frac{1}{2^3} <\sum_{n=1}^\infty\frac{1}{n^3} = 1+\sum_{n=2}^\infty \frac{1}{n^3}<1+\sum_{n=2}^\infty\frac{1}{n^3-n} &= 1+\sum_{n=2}^\infty\frac{1}{n(n-1)(n+1)} \\ &= 1+\frac{1}{2}\sum_{n=2}^\infty\left(\frac{1}{n(n-1)}-\frac{1}{n(n+1)}\right) \\ &= 1 + \frac{1}{2}\lim_{k\to\infty}\left(\frac{1}{2}-\frac{1}{k(k+1)}\right) \\ &= \frac{5}{4}.\end{align}
