# Prove that ${e\over {\pi}}\lt{1\over {2\gamma}}$ without using a calculator.

I started the proving with the verification of ${\gamma}\lt{1\over\sqrt3}$ inequality (without using calculator). If it is proved then my question is also proved. For this I used the ${e\over {\pi}}\lt{\sqrt3\over{2}}$ inequality (Prove that ${e\over {\pi}}\lt{\sqrt3\over{2}}$ without using a calculator.)

So I performed an approximation of ${1\over\sqrt3}$ with help of ${f(x)=x^2-3}$ using Newton method. Accordance with it ${1\over\sqrt3}\gt{10864\over18817}$.

On the other hand I am going to use the expansion of ${\gamma}$ constant is due to Fontana and Mascheroni. But there should be summarize large number of Gregory coefficients to reach the required accuracy, I am rather looking for better approximation.

I need help to finalize my solution and/or proved my question.

• I think it's impossible. – Michael Rozenberg Oct 13 '17 at 14:48
• @MichaelRozenberg I agree , unless we calculate all the numbers precise enough. – Peter Oct 13 '17 at 15:04
• You can check how many decimal places of accuracy you need on each constant, for the relation to hold (e.g. $\frac{e}{\pi}\not<\frac{1}{2\times 0.58},\frac{1}{2\times 0.578}$ but $\frac{e}{\pi}<\frac{1}{2\times 0.5773}$). So this would seem to imply that we need to know $\gamma$ to at least $4$ dp., $\pi$ to at least $2$ dp. and $e$ to at least $3$ dp. So to do this without calculating the expansions of the constants seems like a bit of a big ask. – Jam Oct 13 '17 at 15:27
• – Jack D'Aurizio Oct 13 '17 at 16:39

## 3 Answers

You could try a Taylor series argument. Take $exp(x)=\sum_{n=1}^{\infty}\frac{x^n}{n!}$. You plug in $\frac{1}{\pi}$ for $x$. Since $exp(x)$ is analytic everywhere (i.e. the Taylor series converges $\forall x\in \mathbb{R}$), We now that $\sum_{n=1}^{\infty}\frac{(\frac{1}{\pi})}{n!}=\frac{e}{\pi}$ as $n\rightarrow \infty$. We can then compare it to a suitable series involving $\gamma$. We need to have a series where $S_N\rightarrow \frac{1}{2 \gamma}$ as $N\rightarrow \infty$.

I suppose you could try to find a Taylor expansion for $1/x$ but if I recall Taylor series of rational functions are tedious to work with. However, in principle, this will work. Note that since we are not expanding about $0$, we are ok, since rational functions are analytic away from singularities. We can then show all the partial sums of the exponential Taylor series are strictly less than the partial sums of the series for $\frac{1}{2 \gamma}$ (or at least show the exponential series is smaller as $N\rightarrow \infty$). This technique is easily generalizable, the difficulty arises in finding suitable series to compare to each other.

• thank you for the explanation, I am going to try the method. – JV.Stalker Oct 13 '17 at 16:30

## Upper bound of $e$

$$\frac{1}{7!}\approx0.000198\ldots<0.000457\ldots\frac{1}{3^7}$$

Therefore, for $n\geq7$, $\frac{1}{n!}<\frac{1}{3^n}$

\begin{aligned} e &=\sum_{0\leq n\leq7}\frac{1}{n!}+\sum_{8\leq n}\frac{1}{n!} \\e&<\sum_{0\leq n\leq 7}\frac{1}{n!}+\sum_{n=8}^\infty\frac{1}{3^n} \\e&<\sum_{0\leq n\leq 7}\frac{1}{n!}+\frac{1}{3^8}\sum_{n=0}^\infty\frac{1}{3^n} \\e&<2.71825\ldots+\frac{1}{3^8}\left(\frac{1}{1-1/3}\right) \\e&<2.71848 \end{aligned}

## Upper bound of $\gamma$

This part's tricky since we need a relatively high precision (at least $4$ decimals) for $\gamma$ but it's also quite hard to bound from above. It also appears that we can't escape long computations at this stage, but they're well within the remit of someone to do by hand, in a day. The computations I've chosen requires finding binomial coefficients (e.g. from Pascal's triangle), finding their reciprocals, and summing around $200$ numbers. I think this should be feasible without a calculator, even with just a couple of hours. Euler managed to do it in the $1700$s so surely we can too!

Let $S(k)=\sum_{0\leq j\leq k-1}\binom{2^{k-j}+j}{j}^{-1}$. As $k\to\infty$, $S(k)$ appears to approach $1$ from above, though I haven't found a proof of this yet.

$S(20)<1.004902$ so for $k\geq20$, $S(k)<1.004902$.

\begin{aligned} \gamma &=\sum_{1\leq k}\frac{S(k)}{2^k} \\\gamma&<\sum_{1\leq k\leq20}\frac{S(k)}{2^k}+\sum_{20< k}\frac{1.004902}{2^k} \\\gamma&<\sum_{1\leq k\leq20}\frac{S(k)}{2^k}+1.004902\sum_{k=21}^\infty\frac{1}{2^k} \\\gamma&<\sum_{1\leq k\leq20}\frac{S(k)}{2^k}+\frac{1.004902}{2^{21}}\sum_{k=0}^\infty\frac{1}{2^k} \\\gamma&<\sum_{1\leq k\leq20}\frac{S(k)}{2^k}+\frac{1.004902}{2^{21}}\left(\frac{1}{1-1/2}\right) \\\gamma&<0.57721470\ldots+0.000000958\ldots \\\gamma&<0.57721567 \end{aligned}

## Lower bound of $\pi$ and Final Result

By Archimedes, $\pi>3.14$ so we can show $$2\gamma e<2\cdot2.71848\cdot0.57721567<3.14<\pi$$

Hence, $\frac{e}{\pi}<\frac{1}{2\gamma}$, as desired.

We could also find a lower bound for $\pi$ by truncating an alternating series. But many such series require vast numbers of terms to even get the $2$ decimal places that we need. For example, the series $\frac{2}{\pi} =\sum_{0\leq k}\frac{(-1)^k(4k+1)((2k-1)!!)^3}{((2k)!!)^3}$, due to Ramanujan still takes a few hundred terms to reach a lower bound of $3.14$, which is far too long to do by hand. It was tricky finding a way of bounding $\gamma$ since $\gamma =\sum_{1\leq n}\frac{(-1)^n\log_2{n}}{n}$ and $\gamma<\int_1^{N}\left(\frac{1}{\lfloor x \rfloor}-\frac{1}{x}\right)\,\mathrm{d}x+\int_{N}^{+\infty}\frac{\mathrm{d}x}{x(x-1)}$ both converge too slowly.

• You chose a famous but not useful (here) series of Ramanujan. For lower bound of $\pi$ you can use Ramanujan's (not so) famous series $$\frac{4}{\pi}=\frac{1123}{882}-\frac{22583}{882^{3}}\cdot\frac{1}{2}\cdot\frac{1\cdot 3}{4^{2}}+\frac{44043}{882^{5}}\cdot\frac{1\cdot 3}{2\cdot 4}\cdot\frac{1\cdot 3\cdot 5\cdot 7}{4^{2}\cdot 8^{2}}-\dots$$ and the first term alone gives $\pi>4\times 882/1123=3.1415850\dots$ +1 for the nice answer. – Paramanand Singh Oct 23 '17 at 6:21
• You may be interested in this question (math.stackexchange.com/q/2485558/72031) which is inspired from this answer. – Paramanand Singh Oct 23 '17 at 7:38
• @ParamanandSingh Thanks for the suggestion and the +1 :) I was considering some other series by Ramanujan which converge pretty rapidly but imo it begs the question of whether they're really done without a calculator since (as far as I know) it's hard to prove some of Ramanujan's formulas without long calculations (I might be wrong about that though). Thanks for that link too, it's quite interesting :) – Jam Oct 23 '17 at 14:36
• Ramanujan derived these series via hand calculation using his super powers of algebraic manipulation. Later authors proved his results using a combination of theory and machine calculation and this series was also proved with the help of machine calculation. So you are right in a sense. – Paramanand Singh Oct 23 '17 at 18:46
• @ParamanandSingh Ah, I see; that makes sense. I figured he must have done it by hand since it's unlikely he'd have had a calculator. I think it's an interesting idea though: whether we're allowed to use theorems "without a calculator" (even if the theorem itself was like the 4 color theorem and proven with one) – Jam Oct 23 '17 at 19:39

I have an interesting approach. The Shafer-Fink inequality and its generalization allow to devise algebraic approximations of the arctangent function with an arbitrary uniform accuracy. By a change of variable, the same holds for the hyperbolic arctangent function over the interval $(0,1)$ and for the logarithm function over the same interval. For instance,

$$\forall x\in(0,1),\qquad \log(x)\approx\frac{90(x-1)}{7(x+1)+12\sqrt{x}+32\sqrt{2x+(x+1)\sqrt{x}}}\tag{A}$$ and $\approx$ holds as a $\leq$, actually. We have $$\gamma = \int_{0}^{1}-\log(-\log x)\,dx \tag{B}$$ hence: $$\gamma\leq 1-\log(90)+\int_{0}^{1}\log\left[7(x+1)+12\sqrt{x}+32\sqrt{2x+(x+1)\sqrt{x}}\right]\,dx\tag{C}$$ where the RHS of $(C)$ just depends on the (logarithms of the) roots of $7 + 32 x + 12 x^2 + 32 x^3 + 7 x^4$, which is a quartic and palindromic polynomial.
The numerical approximation produced by $(C)$ allows to state: $$\gamma < 0.5773534 < \frac{\pi}{2e}.\tag{D}$$ Actually $(A)$ is not powerful enough to prove $\gamma<\frac{1}{\sqrt{3}}$, but we can achieve that too by replacing $(A)$ with the higher-order (generalized) Shafer-Fink approximation.

• thanks for the elegant solution I have to study this. – JV.Stalker Oct 13 '17 at 18:40