Show that:


I have tried power series of exponential around $0$ wich is :


We can pursue it with another order .

Edit :

An inequality due to Nanjundiah states for $n\geq 1$ a natural number:


Edit 2 :

An inequality due to Bennett states for $x$ a real number $m,n$ natural numbers and $m,n>x$ then :


There is a little mistake in the statement of the edit 2 we need $x\neq 0$

For $\pi$ I have tried the continued fraction see wikipedia

Edit 3 :

I transform the product into a sum using :

$$\ln(x)\leq x-1$$

For $x>0$ so we need to show :


Wich is true numerically but I have not a proof yet of this transformation .

A interesting function is :


Numerically we have $f(x)<0$ for $x>0$ and $f'(1)=0$

Question: How to show it by hand without a calculator?

  • 1
    $\begingroup$ This is very close to $1$. Why do you expect that this can be decided by hand ? $\endgroup$
    – Peter
    Commented Dec 12, 2021 at 13:17
  • 1
    $\begingroup$ @Peter It's more a challenge and I continue to hope ^^. $\endgroup$ Commented Dec 12, 2021 at 13:48
  • 1
    $\begingroup$ @ErikSatie Simplify with $$\frac{\sqrt \frac e2(e^2-2)}{2\pi}$$ $\endgroup$ Commented Dec 12, 2021 at 14:32
  • 1
    $\begingroup$ My question : how do you make so nice problems ? $\endgroup$ Commented Dec 15, 2021 at 7:53
  • 1
    $\begingroup$ I tried to 'reverse engineer it' by viewing $1/\sqrt{2}(1/2 e^{5/2}-e^{1/2})$ as the result of integrating some function of $r$ over $(0,1)$ and then trying to show that function is bounded by $1$; this would do it, as one could the inequality follows by integrating over the unit circle. Nothing yet but I'll keep trying. $\endgroup$
    – Integrand
    Commented Dec 15, 2021 at 21:46

5 Answers 5


A piece of sheer numerical good luck makes the calculation fairly painless.

I'll take this inequality to be well known: $$ \pi > \frac{333}{106}. $$ (I don't know a simple proof. For some that aren't simple, see Is there an integral that proves $\pi > 333/106$?.)

From $e$ Continued Fraction --- from Wolfram MathWorld, we have: $$ e < \frac{193}{71}. $$

Therefore, it is enough to prove: $$ \sqrt{\frac{193}{2\cdot71}} < \frac{333}{106}\left(\frac{193^2}{2\cdot71^2} - 1\right)^{-1}. $$ But \begin{multline*} 193^2 - 2\cdot71^2 = (200 - 7)^2 - 2(70 + 1)^2 = (40000 - 2800 + 49) - 2(4900 + 140 + 1) \\ = 37249 - 10082 = 27167 = 7\cdot3881, \end{multline*} and the right hand side of the required inequality simplifies to: $$ \frac{2\cdot3\cdot111\cdot71^2}{2\cdot7\cdot53\cdot3881} = \frac{426\cdot7881}{371\cdot7762} > \frac{426\cdot7882}{371\cdot7763} = \frac{426\cdot1126}{371\cdot1109} = \frac{426}{371}\left(1 + \frac{17}{1109}\right), $$ where we have used the fact that if $y > x > 0$ then $\frac{y}{x} > \frac{y + 1}{x + 1}.$

Now comes the piece of luck: $$ \frac{1109}{17} = 65 + \frac4{17} < 65 + \frac14 = \frac{261}4, \text{ whence } 1 + \frac{17}{1109} > 1 + \frac4{261}, $$ whence the right hand side of the required inequality is greater than: $$ \frac{426\cdot265}{371\cdot261} = \frac{(6\cdot71)\cdot(5\cdot53)}{(7\cdot53)\cdot(3\cdot87)} = \frac{710}{609}. $$ We now only have to prove: $$ \frac{193}{2\cdot71} < \left(\frac{710}{609}\right)^2, \text{ i.e., }\ 200\cdot71^3 > 193\cdot609^2. $$ We have already calculated $71^2 = 5{,}041,$ so $71^3 = 352{,}870 + 5{,}041 = 357{,}911,$ and $200\cdot71^3 = 71{,}582{,}200.$ On the other hand, $609^2 = 360{,}000 + 10{,}800 + 81 = 370{,}881,$ therefore $193\cdot609^2 = (200 - 7)\cdot370{,}881 = 74{,}176{,}200 - 2{,}596{,}167 = 71{,}580{,}033.$ This proves the inequality.



By excess rational approximations of $e$ can be found from the usual Taylor expansion.

By default rational approximations of $\pi$ can be found from the Machin formula, among others.

Five exact decimals should be enough.

Now the following inequality can be established by hand (though this will take several hours):


(With $\overline e=2.71829$ and $\underline\pi=3.14159$, you get $7.99888\dots<8$.)

Not glamorous, but effective.

  • $\begingroup$ So does interval arithmetic with $e \in [2.7182,2.7183$ and $\pi \in [3.1415,3.1416]$. $\endgroup$
    – lhf
    Commented Dec 15, 2021 at 16:04
  • 2
    $\begingroup$ Several hours is a bit exaggerate; more like 10 to 15 minutes. $\endgroup$
    – Servaes
    Commented Dec 15, 2021 at 16:50
  • $\begingroup$ @Servaes: no, I am assuming starting from scratch and working with fractions only. $\endgroup$
    – user1003895
    Commented Dec 16, 2021 at 9:17
  • $\begingroup$ @lhf: that's right, four decimals are enough, $7.9995<8$. $\endgroup$
    – user1003895
    Commented Dec 16, 2021 at 9:42
  • $\begingroup$ @Guste So am I. I guess it depends on ones experience doing arithmetic. $\endgroup$
    – Servaes
    Commented Dec 16, 2021 at 12:22

Using values rounded up to the fourth decimal,


and down


This takes four multiplies of $4$-decimal numbers and a little more.

(If using precomputed tables is allowed, this one supplies $\sqrt2,\pi,\pi^2,\sqrt e,e,e^2$: http://www.ebyte.it/library/educards/constants/MathConstants.html)

  • 1
    $\begingroup$ Maybe combine answers? $\endgroup$ Commented Dec 16, 2021 at 21:43

I Do not have a very satisfying answer. However I will share my attempt, it might be useful to someone, perhaps the person who, ultimately gives a more sophisticated answer. First, \begin{align} &\sqrt{\frac{e}{2}}\left(\frac{e}{\pi}\right)\left(\frac{e}{2} - \frac{1}{e}\right) < 1, \\[1em] \implies &\frac{\sqrt{e}}{\pi}\left(\frac{e^2 - 2}{2\sqrt{2}}\right) < 1, \\[1em] \implies &e(e^2 - 2)^2 < 8\pi^2. \end{align}

Lower bound on $\pi$:

For the lower bound on $\pi$ consider the following Ramanujan series, \begin{align} \frac{4}{\pi} = \frac{1123}{882} - \frac{22583}{(882)^3}\cdot\frac{1}{2}\cdot\frac{1\cdot 3}{4^2} + \frac{44093}{(882)^5}\cdot\frac{1\cdot 3}{2\cdot 4}\frac{1\cdot 3\cdot 5\cdot 7}{4^2\cdot 8^2} - \dots. \end{align} Therefore, \begin{align} \pi > 4\frac{882}{1123} = \frac{3528}{1123}. \end{align}

Upper bound on $e$:

Now here is where it becomes unsatisfying. The nature of the problem requires a very tight bound on $e$. The first $10$ terms of the Maclaurin series will do, i.e., \begin{align} e &= \sum_{n=0}^{\infty}\frac{1}{n!}, \\ &= \sum_{n=0}^{9}\frac{1}{n!} + \sum_{n=10}^{\infty}\frac{1}{n!}, \\ &\stackrel{(i)}{<} \sum_{n=0}^{9}\frac{1}{n!} + \sum_{n=10}^{\infty}\frac{1}{3^n},\\ &= \sum_{n=0}^{9}\frac{1}{n!} + \frac{1}{3^{10}}\sum_{n=0}^{\infty}\frac{1}{3^n},\\ &= \sum_{n=0}^{9}\frac{1}{n!} + \frac{1}{3^{10}}\left(\frac{1}{1 - \frac{1}{3}}\right),\\ &= \sum_{n=0}^{9}\frac{1}{n!} + \frac{1}{2\cdot3^{9}}. \end{align}

Here $(i)$ holds true since for $n\geq 7$, $\frac{1}{n!} < \frac{1}{3^n}$. Now the finite sum can be written out as, \begin{align} \sum_{n=0}^{9}\frac{1}{n!} &= 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \frac{1}{720} + \frac{1}{5040} + \frac{1}{40320} + \frac{1}{362880}, \\ &= 1 + 1 + \frac{181440}{362880} + \frac{60480}{362880} + \frac{15120}{362880} + \frac{3024}{363880} + \frac{504}{362880} + \frac{72}{362880} + \frac{9}{362880} + \frac{1}{362880},\\ &= \frac{986410}{362880} = \frac{98641}{36288}. \end{align} Taken together with $\tfrac{1}{2\cdot3^9}$ it 'simplifies' to, \begin{align} e < \frac{98641}{36288} + \frac{1}{39366} = \frac{98641}{162\cdot 224} + \frac{1}{162\cdot 243} = \frac{98641\cdot 243}{162\cdot 224 \cdot 243} + \frac{224}{162\cdot 224 \cdot 243} = \frac{a}{b}, \end{align} where $a = 23969987$ and $b = 8817984$, if I am not mistaken.

Now, it is verifiable, by hand, in finite time that, \begin{align} e(e^2-2)^2 < \left(\frac{a}{b}\right)\left(\left(\frac{a}{b}\right)^2 - 2\right)^2 < 8\left( \frac{3528}{1123}\right)^2 < 8\pi^2. \end{align}

However, I do not know how long it would take me. I am curious, whether there exists a more sophisticated answer.

  • 1
    $\begingroup$ Done by hand ?? $\endgroup$
    – user1003895
    Commented Dec 16, 2021 at 9:20
  • $\begingroup$ "one year, one decade or one lifetime": the computation involves integers of length below $40$, what's the problem ? And computing with decimal numbers is even faster. If you take the first decimals for known (as there is no doubt they can be obtained by hand), the computation reduces to four multiplies of five-decimals numbers. A shortcut is probably possible with series for $\pi^2$ or $\pi^{-2}$. $\endgroup$
    – user1003895
    Commented Dec 16, 2021 at 9:32
  • $\begingroup$ @Guste you are right. These calculations are’nt as bad as I initially thought. $\endgroup$
    – vshas
    Commented Dec 16, 2021 at 9:41
  • $\begingroup$ As pointed by @lhf, $e<\frac{27183}{10000}$ and $\frac{31415}{10000}<\pi$ will do the job. In integers, $7894756088154938236143<7895217800000000000000$, but there is no need to compute to full accuracy. $\endgroup$
    – user1003895
    Commented Dec 16, 2021 at 9:54

The simplification of the givens problem product is


This is much easier to be estimated against one.

It goes even more nicely:


As one can see immediately from this representation is that the factor $e+\sqrt{2}$ is the biggest. Since $e \approx 2.7$ and $\sqrt{2} \approx 1.4 $.

So the biggest factor of the nominator is about $4.1325$ and the other two are smaller and than $1.5$ for sure. That is indeed smaller than $2\pi$ that is somewhat bigger than $6$.

So this is multiplying $1.166$ with $1.30$ and $4.13$ and that is smaller than $2\pi$.

So this is a problem connection two real numbers well known and famous with $2\pi$. As well as famous and irrational.

We already used the calculator as well.

To make the result with the calculator more irresistible. The true values are

$6.28268 $ and $6.28319$

This is very little compared to the rounding of the three single factors.

This shows too representation matters! How to get something representing $e$ and $\sqrt{2}$ for simplification?

I hope You are able to follow so far.

A nice repository to expressing $e$ in term of $2$ or $\sqrt{2}$ are well known identities and inequality from the history of Mathematics:


It is now only a matter of how close You intend to approximate the two numbers in comparison to $2\pi$. A shown by the use of the calculator in my representation of the given inequality only necessary up to the third digit after the decimal colon. That is not little, that is not much.

So take for example the infinite product representations. Perhaps that from Pippengers. Or approximate with the limited sequences for $e$, perhaps Sterling's formulas. It is a matter of taste and accuracy and ethics to the numerical Mathematician what he prefers best.

Whatever you select there is always the last step in the accuracy of ease of the equation that will make use of the calculator. Numbers never lie but the paths to get good or even best numerics representations are up to the Mathematician himself.

$\frac{\sqrt{2 e}(\frac{e}{\sqrt{2}}-1)(\frac{e}{\sqrt{2}}+1)}{2\pi}$


Pippengers product:

$e=2\left({\frac {2}{1}}\right)^{{1/2}}\left({\frac {2}{3}}{\frac {4}{3}}\right)^{{1/4}}\left({\frac {4}{5}}{\frac {6}{5}}{\frac {6}{7}}{\frac {8}{7}}\right)^{{1/8}}\cdots $

Put this into the given nominator and yeah calculator or round in the exactness of the formulare. These are just infinite products. Look at Infinite product.

You might make use of Wallis product to approximate $2\pi$.

Some are fine with multiplicating some prefer to sum the bridge is taking logarithms.

This inequality in numbers is


That is evidently true.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .