Proof that $\frac{9}{8} < \sum\limits_{n=1}^{\infty} \frac{1}{n^3} < \frac{5}{4}$.

Question

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.

Answer 1

$$\sum_{n=1}^{\infty} \frac{1}{n^3} =\sum_{n=1}^k \frac{1}{n^3} + \sum_{n=k+1}^{\infty} \frac{1}{n^3}< \sum_{n=1}^k \frac{1}{n^3}+\int_{k}^{\infty} \frac{1}{x^3} dx =\sum_{n=1}^k \frac{1}{n^3}+\frac{1}{2 k^2}$$ and this for any $k\ge 1$. For $k=2$ we get RHS $=1+1/8+1/8=5/4$.

Answer 2

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.

Answer 3

See green bars for LH inequality and orange bars for RH inequality. $$\begin{align} 1+\int_2^\infty\frac 1{x^3}\; dx \quad &<\quad \sum_{n=1}^\infty \frac 1{n^3} &&<\quad 1+\frac 18+\int_2^\infty \frac 1{x^3} \; dx\\ \frac 98 \quad &<\quad \sum_{n=1}^\infty \frac 1{n^3} &&<\quad\frac 54 & \end{align}$$

Answer 4

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 $$ \frac98\lt\sum_{n=1}^\infty\frac1{n^3}\lt\frac{11}9 $$

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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, $$ \frac98\le\sum_{n=1}^\infty\frac1{n^3}\le\frac54 $$

Answer 5

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}