Proof that $0< e-\ln 2-2 < \frac 3 {100}$

Question

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.

Answer 1

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

Answer 2

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 $$e\color{red}{-\ln2}-2<1+1+\frac12+\frac16+\frac1{24}+\frac1{120}+\frac2{720}\color{red}{-\frac12-\frac18-\frac1{24}-\frac1{64}-\frac1{180}-\frac1{384}}-2=\frac5{192}+\frac1{5760}<0.027$$

Answer 3

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

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

Answer 4

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.