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You may have learnt the estimates $e\approx 2.7$ and $\ln 2 \approx 0.7$, thus yielding $e-\ln 2 \approx 2$. A calculator indicates more precisely that $e-\ln 2\approx 2.025$.
Out of curiosity I'd like to see self-contained (requiring only pen and paper) proofs that $$0< e-\ln 2-2 < \frac 3 {100}.$$

Here's my attempt. Surely by power series expansion, $$e-\ln 2-2 = -\frac 16+\sum_{k=4}^\infty (-1)^k\left(\frac 1{k}+\frac{(-1)^k}{k!}\right),$$ and it is easily verified that the series in the RHS is alternating. Consequently we obtain the bound $a_n \leq e-\ln 2-2 \leq b_n$ for each $n\geq 4$, where $$a_n = -\frac 16 +\sum_{k=4}^{2n+1} (-1)^k\left(\frac 1{k}+\frac{(-1)^k}{k!}\right)$$ and $$b_n = -\frac 16 +\sum_{k=4}^{2n} (-1)^k\left(\frac 1{k}+\frac{(-1)^k}{k!}\right).$$

A computer indicates that $a_{10}>0$ and $b_{52}<\frac 3 {100}$, thus yielding a proof, which is not satisfactory since it requires a computer/calculator.

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  • $\begingroup$ You need 160 terms of the series for $\ln 2$ to conclude that $ln 2 > 0.69$. $\endgroup$ Commented Oct 2, 2024 at 15:20
  • $\begingroup$ @lhf I'd like to see a proof that requires only pen and paper. $\endgroup$ Commented Oct 2, 2024 at 15:21
  • $\begingroup$ Why the downvote ? $\endgroup$ Commented Oct 2, 2024 at 15:36
  • $\begingroup$ Here is a fast series for $\ln 2$: math.stackexchange.com/a/61283/589 $\endgroup$ Commented Oct 2, 2024 at 16:18
  • $\begingroup$ Because $100\ln2>69$ $\endgroup$ Commented Oct 2, 2024 at 17:20

4 Answers 4

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we have $$\int_0^1 x^2(1-x)^2(a+bx)e^xdx=2(7a-32b)e+2(-19a+87b)=0.03-e+\ln 2+2>0$$ since $a> 0$ and $a+b>0$. (Values here)

Edit: $$\int_0^1 \frac{x^3(1-x)^3(a+bx)}{1+x}dx=a\left(\frac{111}{20}-8\ln 2\right)+b\left(8\ln 2-\frac{194}{35}\right)=\ln 2-\frac{551}{800}>0$$ Values here

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  • $\begingroup$ The system of equations in the variables $a,b$ is $3b-a=-1$, $3a-8b=\ln 2 + 2 + 3/100$. The solutions are $a=3\ln(2)-1.91\approx 0.16$ and $b=0.97+\ln 2\approx -0.27$. Thus $a+b<0$. $\endgroup$ Commented Oct 2, 2024 at 15:42
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    $\begingroup$ Where did this integral come from? $\endgroup$ Commented Oct 2, 2024 at 16:15
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    $\begingroup$ @BoweiTang I agree with your calculations. However you need to show that both $a$ and $b$ are positive, which amounts to proving that $\ln 2 > \frac{551}{800}$. This inequality is quite sharp, do you know a proof ? $\endgroup$ Commented Oct 2, 2024 at 16:19
  • $\begingroup$ Doesn't this solution need 2-digit approximations of $\ln 2$? Just like mine? $\endgroup$ Commented Oct 2, 2024 at 16:19
  • $\begingroup$ @Marloumarlou Updated. Please check this one. $\endgroup$ Commented Oct 2, 2024 at 16:43
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Using $$\ln 2=\sum_{n=1}^\infty\frac1{n2^n}$$ and $$e<\sum_{n=0}^N\frac1{n!}+\frac2{(N+1)!}\;\text{for}\;N>3$$ We have $$e\color{red}{-\ln2}-2<1+1+\frac12+\frac16+\frac1{24}+\frac1{120}+\frac2{720}\color{red}{-\frac12-\frac18-\frac1{24}-\frac1{64}-\frac1{180}-\frac1{384}}-2=\frac5{192}+\frac1{5760}<0.027$$

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We can obtain the result using that, by few terms

  • $e-2 \approx \frac 1 2+\frac 1{3!}+\frac1{4!}+\frac1{5!}+\frac1{6!}\approx 0.718$

and using inverse hyperbolic tangent series estimation

$$\ln (n+1) = \ln(n) + 2\sum_{k=0}^\infty\frac{1}{2k+1}\left(\frac{1}{2 n+1}\right)^{2k+1}$$

for $n=1$ with just three terms we can estimate

  • $\log 2\approx \frac23+\frac2{81}+\frac2{1251} \approx0.693 $

and therefore

$$e-\ln 2-2 \approx 0.718-0.693 =0.025$$

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    $\begingroup$ What an ugly series... +1 $\endgroup$ Commented Oct 2, 2024 at 18:57
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We can prove the claim directly using only the Maclaurin series for $\ln(1+u)$.

I recall answering this before, but can't find the reference, so I rederive. To wit, putting $u\to y$ and then $u\to -y$ into the Maclaurin series gives

$\ln(1+y)=y-\dfrac{y^2}2+\dfrac{y^3}3-\dfrac{y^4}5+\dfrac{y^5}5-\ldots$

$\ln(1-y)=-y-\dfrac{y^2}2-\dfrac{y^3}3-\dfrac{y^4}5-\dfrac{y^5}5-\ldots$

We subtract the second series from the first and express the left side as the logarithm of a quotient:

$\ln\left(\dfrac{1+y}{1-y}\right)=2y+\dfrac{2y^3}3+\dfrac{2y^5}5+\ldots$

and finally put $y=(x-1)/(x+1)$ to reduce the left side to $\ln x$:

$\ln(x)=2\left(\dfrac{x-1}{x+1}\right)+\dfrac23\left(\dfrac{x-1}{x+1}\right)^3+\dfrac25\left(\dfrac{x-1}{x+1}\right)^5+\ldots\tag{1}$

where the series converges whever $|x-1|<|x+1|$ (meaning all positive $x$) and has all terms positive when $x>1$.

Now using the terms shown we put $x=2$ into (1) and obtain

$\ln(2)>\dfrac23+\dfrac23\left(\dfrac13\right)^3+\dfrac25\left(\dfrac13\right)^5>0.693.$

The figure $0.693$ is correct to five decimal places (excludingvthe rest of the terms in the series) using a standard calculator.

The claim is then proved if $e<2.03+0.693=2.723$. So we put $x=2.723$ into $(1)$ and observe that the three terms shown in logarithmic series already give a natural logarithm greater than $1$:

$\ln(2.723)>2\left(\dfrac{1.723}{3.723}\right)+\dfrac23\left(\dfrac{1.723}{3.723}\right)^3+\dfrac25\left(\dfrac{1.723}{3.723}\right)^5>1.00017,$

where we again take five places after the decimal point from a calculator result.

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