How do I show that the sum of the reciprocals of perfect cubes is less than 3/2?

Question

I have done research to find that this value is known as Apery's constant, but I haven't found anything about calculating it using purely math. I have tried to find patterns between the numbers but have come up with nothing. Please help and explain in simple terms, using solely math.

Answer 1

As you've mentioned, Apery's constant is $\zeta(3) = \sum_n \frac{1}{n^3}$, where of course $\zeta(s) = \sum_n \frac{1}{n^s}$ is the Riemann Zeta Function.

Let's write the first few terms of this sum:

$$ \frac{1}{1^3} + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + \frac{1}{5^3} + \frac{1}{6^3} + \frac{1}{7^3} + \frac{1}{8^3} + \frac{1}{9^3} + \frac{1}{10^3} + \ldots $$

To estimate this we'll use a common trick: upper bound $\frac{1}{n^3}$ by $\frac{1}{(2^k)^3}$ where $2^k$ is the power of $2$ immediately before $n$. (Do you see why this is an upper bound?)

Now let's apply this upper bound to each term of the above sum:

$$ \frac{1}{1^3} + \frac{1}{2^3} + \frac{1}{2^3} + \frac{1}{4^3} + \frac{1}{4^3} + \frac{1}{4^3} + \frac{1}{4^3} + \frac{1}{8^3} + \frac{1}{8^3} + \frac{1}{8^3} + \ldots $$

Do you see the pattern here? Collecting terms, we see that

$$ \begin{align} \sum_n \frac{1}{n^3} &= \frac{1}{1^3} + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + \frac{1}{5^3} + \frac{1}{6^3} + \frac{1}{7^3} + \frac{1}{8^3} + \frac{1}{9^3} + \frac{1}{10^3} + \ldots \\ &\leq \frac{1}{1^3} + \frac{1}{2^3} + \frac{1}{2^3} + \frac{1}{4^3} + \frac{1}{4^3} + \frac{1}{4^3} + \frac{1}{4^3} + \frac{1}{8^3} + \frac{1}{8^3} + \frac{1}{8^3} + \ldots \\ &= \frac{1}{1^3} + \frac{2}{2^3} + \frac{4}{4^3} + \frac{8}{8^3} + \ldots \\ &= \sum_k \frac{1}{(2^k)^2} \\ &= \sum_k \frac{1}{4^k} \end{align} $$

Of course, we recognize $\sum_k \frac{1}{4^k}$ as a geometric series whose sum is $\frac{1}{1- \frac{1}{4}}$. So at the end of the day, we learn

$$ \zeta(3) = \sum_n \frac{1}{n^3} \leq \frac{1}{1 - \frac{1}{4}} = \frac{4}{3} < \frac{3}{2} $$

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I hope this helps ^_^

Answer 2

Since $f(x) = \frac1{x^3}$ is a decreasing function, its right-hand Riemann sums are less than the corresponding integrals; therefore $$ \sum_{n=1}^\infty \frac1{n^3} = 1 + \frac18 + \sum_{n=3}^\infty \frac1{n^3} < 1 + \frac18 + \int_2^\infty \frac1{x^3} \,dx = \frac54, $$ which is certainly less than $\frac32$.

Answer 3

The other answers give you tighter estimates, I think, but since your upper bound is quite relaxed (the sum you're looking for is actually around 1.2) let's have some fun with the fact that $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$$

Since ${n^3} \ge {n^2}$ (with equality only for $n=1$),

$$\sum_{n=1}^\infty \frac{1}{n^3} < \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$$

That's still above 1.5, but you only need to correct a few terms to get to where you want to be:

$$\sum_{n=1}^\infty \frac{1}{n^3} < \frac{\pi^2}{6} + \left( \frac{1}{2^3} - \frac{1}{2^2} \right) + \left( \frac{1}{3^3} - \frac{1}{3^2} \right)$$

Let's be lazy and estimate $\pi^2 < 10$, then you end up with 317 / 216 which is smaller than 3 / 2 = 324 / 216.

Answer 4

Luckily, there is a formula for the Riemann zeta function automatically implies that $\zeta(3) < \frac{3}{2}$.

$$ \zeta(s) = \frac{s}{s-1} - s\int_{1}^{\infty}\frac{x - [x]}{x^{s+1}}dx $$

where $[x]$ is the floor function. For $s = 3$, the integral is positive hence

$$ \zeta(3) = \frac{3}{2} - 3\int_{1}^{\infty}\frac{x - [x]}{x^{4}}dx < \frac{3}{2} $$