Prove that $\frac{7}{12}<\ln 2<\frac{5}{6}$ using real analysis I studying in Real Analysis 2, but I have no idea how to solve this problem. My guess is to use Mean Value Theorem or a similar theorem? Could any one help me?
Thanks.
 A: To the extent that arithmetic is part of real analysis, it suffices to note that 
$$3^7=2187\lt4096=2^{12}$$
and
$$2^6\cdot4^5=2^{16}=65536\lt100000=10^5=2.5^5\cdot4^5$$
The desired inequalities now follow from the fact that $2.5\lt e\lt3$.
A: J.G.'s elegant answer deserves to be restated in the following simple (and perhaps more familiar) form: we can approximate the integral
$$\ln 2 = \int_2^4 \frac1x\, dx$$
by a Riemann sum splitting the interval [2,4] into two parts of width $1$.  Since $1/x$ is strictly decreasing on this interval, the left-Riemann sum is a (strict) overestimate, and the right-Riemann sum is a (strict) underestimate.  Thus:
$$\frac{7}{12} = \frac13 + \frac14 < \int_2^4 \frac1x\, dx < \frac12 + \frac13 = \frac56.$$
A: There is a wealth of series which converge to $\log(2)$. One which converges rapidly, is
\begin{equation}
\log(2) = \sum_{k=1}^\infty \frac{1}{k 2^k}.
\end{equation}
This is a sum which is closely related to the geometric series and its anti-derivative. Once you have derived this sum, your estimates are sure to follow.
A: One series for $\log(2)$ is
$$
\log(2)=1-\frac12+\frac13-\frac14+\dots
$$
For an alternating series, the terms of which are decreasing in absolute value, the sum is between any two consecutive partial sums, such as
$$
1-\frac12+\frac13=\frac56
$$
and
$$
1-\frac12+\frac13-\frac14=\frac7{12}
$$
A: We wish to prove $H_4-H_2<\ln 2<H_3-H_1$ with $$H_n:=\sum_{k=1}^n\tfrac{1}{k}=\sum_{k=1}^n\int_{k}^{k+1}\frac{dx}{\left\lfloor x\right\rfloor}$$ so that $$H_{n+2}-H_{n}=\int_{n+1}^{n+3}\frac{dx}{\left\lfloor x\right\rfloor}\in\left(\int_{n+1}^{n+3}\frac{dx}{x},\,\int_{n+1}^{n+3}\frac{dx}{x-1}\right)=\left(\ln\frac{n+3}{n+1},\,\ln\frac{n+2}{n}\right).$$ Substituting $n=1$ gives $H_3-H_1 > \ln 2$; substituting $n=2$ gives $H_4-H_2<\ln 2$.
A: Consider the following inequalities from Proof that $\frac{2}{3} < \log(2) < \frac{7}{10}$
as suggested by @labbhattacharjee
$$\frac{1}{2}\int_{0}^{1}\frac{x^2(1-x)}{1+x}dx>0$$
$$\frac{1}{2}\int_{0}^{1}\frac{x^5(1-x)}{1+x}dx>0$$
The integrals are positive because the integrands are positive for $0<x<1$.
Let us evaluate the first integral
$$\begin{align}
\frac{1}{2}\int_{0}^{1}\frac{x^2(1-x)}{1+x}dx
&=
\frac{1}{2}\int_{0}^{1}\frac{x^2-x^3}{1+x}dx \\
&=
\frac{1}{2}\int_{0}^{1}\left(-x^2+2x-2+\frac{2}{1+x}\right)dx\\
&=
\left.\frac{1}{2}\left(-\frac{x^3}{3}+x^2-2x+2\log(1+x)\right) \right|_{0}^{1}\\
&=
\frac{1}{2}\left(-\frac{1}{3}+1-2+2\log(2)\right) \\
&=
\log(2)-\frac{2}{3}>0 \\
\end{align}
$$
Similarly,
$$\frac{1}{2}\int_{0}^{1}\frac{x^5(1-x)}{1+x}dx=\frac{7}{10}-\log(2)>0$$
Combining both results,
$$\frac{2}{3}-\log(2)<0<\frac{7}{10}-\log(2)$$
so
$$\frac{2}{3}<\log(2)<\frac{7}{10}$$
To solve your problem we shall compare $\frac{2}{3}$ against $\frac{7}{12}$ and $\frac{7}{10}$ against $\frac{5}{6}$
On the one hand,
$$21<24$$
$$7·3<2·12$$
$$\frac{7}{12}<\frac{2}{3}$$
On the other hand,
$$42<50$$
$$7·6<5·10$$
$$\frac{7}{10}<\frac{5}{6}$$
Finally,
$$\frac{7}{12}<\frac{2}{3}<\log(2)<\frac{7}{10}<\frac{5}{6}$$
so
$$\frac{7}{12}<\log(2)<\frac{5}{6}$$
A: Similarly to the answer by robjohn, regrouping the alternating series separates lower and upper estimates into two monotonic series
$$\sum_{k=0}^\infty\frac{1}{(2k+1)(2k+2)}=\log(2)=1-\sum_{k=0}^\infty\frac{1}{(2k+2)(2k+3)}$$
Truncating yields
$$0<\frac{1}{2}<\frac{7}{12}<\frac{37}{60}<...\log(2)...<\frac{319}{420}<\frac{47}{60}<\frac{5}{6}<1$$
