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


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:


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


One of the approximants will be:

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


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

  • 2
    $\begingroup$ Square your last once more and $\displaystyle \frac{4^8}{3^8} = \frac{65536}{6561}< 10$ is obvious. Not sure why you would want to have a more complex way to show that. $\endgroup$
    – Macavity
    Apr 6, 2016 at 11:26
  • $\begingroup$ You mean $\frac{4^8}{3^8}$. Yes, it's obvious, but the integral would be more interesting $\endgroup$
    – Yuriy S
    Apr 6, 2016 at 11:29
  • $\begingroup$ $\left(\dfrac{4}{3}\right)^4 = 3.16...$, but $\left(\dfrac{4}{3}-\dfrac{1}{500}\right)^4 = 3.1415...$ and $\left(\dfrac{4}{3}-\dfrac{1}{500}\left(1-\dfrac{1}{10^3}\right)\right)^4 = 3.141592...$ $\endgroup$ Apr 23, 2017 at 22:26

2 Answers 2


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, $\pi\approx\frac{22}{7}$.


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






as well.


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


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