Is there integral or series for $\sqrt{10}-\frac{4^4}{3^4}$ (to prove the inequality)?

Question

Both of these numbers are bad approximations for $\pi$, but they turn out to be much closer together:

$$\sqrt{10}-\frac{4^4}{3^4}=0.00178$$

Since there is a lot of questions here about integrals and series which prove such close inequalities, I wanted to know if something exists for this inequality as well:

$$\sqrt{10}>\frac{4^4}{3^4}$$

If we use the continued fraction for $\sqrt{10}$:

$$\sqrt{10}=3+\cfrac{1}{6+\cfrac{1}{6+\cdots}}$$

One of the approximants will be:

$$\sqrt{10} \approx \frac{117}{37}=\frac{234}{74}$$

$$\frac{4^4}{3^4}=\frac{256}{81}$$

This is just a curiosity for me, there is no other context for the question.

Answer 1

A series can be matched to your inequality.

The number $\sqrt{10}$ is $\sqrt{\frac{4^8}{3^8}+\frac{74}{6561}}$, which is $\sqrt{\frac{4^8}{3^8}+x}$ for $x=\frac{74}{6561}$.

The expansion has alternating sign

$\sqrt{\frac{4^8}{3^8}+x} = \frac{256}{81}+\frac{81}{512}x-\frac{531441}{134217728}x^2+...$

and its first term is $\frac{256}{81}=\frac{4^4}{3^4}$, so the other terms are a series expansion for $\sqrt{10}-\frac{4^4}{3^4}$.

Here is a related definite integral that proves the inequality $\sqrt{10}-\frac{256}{81}>0$ because the integrand is positive in $(0,1)$.

$$\int_0^1 \frac{37}{3^4\sqrt{2^{16}+2·37 x}}dx=\sqrt{10}-\frac{256}{81}>0$$

Another possibility is $$\int_0^1 \frac{145624+4095x^2(1-x)^2}{26520048\sqrt{9+x}}dx = \sqrt{10}-\frac{256}{81}$$

Fractions $\frac{234}{74}$ and $\frac{256}{81}$ are related to another notable approximation, $\pi\approx\frac{22}{7}$.

$$\frac{234+22}{74+7}=\frac{256}{81}$$

If you want to think in terms of $\pi^2$ instead of $\pi$ in order to eliminate the root, you can also derive the series

$$10-\left(\frac{256}{81}\right)^2 = \frac{592}{3^8} \sum_{k=0}^\infty \left( \frac{1}{((k+1)(k+2))^3}-\frac{1}{((k+2)(k+3))^3}\right)$$

which also proves your claim because the positive term in the summation is always larger than the negative one (compare their denominators) and

$$10-\left(\frac{256}{81}\right)^2>0$$

is

$$\left(\sqrt{10}+\frac{256}{81}\right)\left(\sqrt{10}-\frac{256}{81}\right)>0,$$

so

$$\sqrt{10}-\frac{256}{81}>0$$

as well.

Answer 2

Well, this is the same as $\sqrt{2}\sqrt{5} - \frac{2^8}{3^4}$. So, you can use tools from calculus to show that this function is positive at x = 2:

$f(x) = \sqrt{x}\sqrt{5} - \frac{x^8}{3^4}$

First of all, its zero is at:

$\frac{x^8}{\sqrt{x}} = 3^4 \sqrt{5} $

This gives us:

$x= 3^{(8/15)} 5^{(1/15)}$

This is slightly larger than 2.

Also, the derivative of f(x) is:

$f'(x) = \frac{\sqrt{5}}{2 \sqrt{x}} - \frac{8x^7}{3^4}$

This is negative for $x > (\frac{3}{2})^{(8/15)} 5^{(1/15)}$ . So, $f(x)$ must be positive before $x= 3^{(8/15)} 5^{(1/15)}$.