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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.

Jaume Oliver Lafont
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Helen
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    Can you show us some of the work you've done? What have you tried with the Mean Value Theorem? How are you going to use the Mean Value Theorem to solve this problem? Showing us some work and giving us something to work off of would really help us. Even if you've tried really hard and haven't gotten anywhere, just show us what you've tried or what you've been thinking of. – Noble Mushtak Feb 07 '16 at 14:19
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    Have you covered alternating series? Those specific estimates to $\ln 2$ suggest to me strongly a certain series... – Jyrki Lahtonen Feb 07 '16 at 14:21
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    To expand on @NobleMushtak's comment, real analysis is very broad, and so there are probably many possible solutions, not all of which might apply to your context. (There's one way to get even tighter bounds using just some elementary integral calculus.) – Travis Willse Feb 07 '16 at 14:21
  • since f is continuous on the closed interval [a, b], where a < b, and differentiable on (a, b), then there exists a point c in (a, b) such that f(b)-f(a)= f'(c) (b-c). so we might could find a boundary at a=7/12 and b=5/6? – Helen Feb 07 '16 at 14:25
  • @Helen Good thinking, but in order to find these boundaries, you need to somehow compute approximations for $e^{\frac{7}{12}}$ and $e^{\frac{5}{6}}$. I would use Taylor's Theorem to help you computer either these values or to compute $\ln 2$. – Noble Mushtak Feb 07 '16 at 14:30
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    @Helen, Related : http://math.stackexchange.com/questions/1642783/proof-that-frac23-log2-frac710 – lab bhattacharjee Feb 07 '16 at 14:44

7 Answers7

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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 $$ 1-\frac12+\frac13-\frac14=\frac7{12} $$

robjohn
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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 $2.5\lt e\lt3$.

Barry Cipra
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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:

$$\frac{7}{12} = \frac13 + \frac14 < \int_2^4 \frac1x\, dx < \frac12 + \frac13 = \frac56.$$

Erick Wong
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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.

Carl Christian
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  • Each partial sum gives a lower bound. However, to get an upper bound, one needs to use something like $$\log(2)\le\sum_{k=1}^n\frac1{k2^k}+\frac1{(n+1)2^n}$$ – robjohn Feb 07 '16 at 16:55
  • Yes, that is true. It is always a balance to determine the proper amount of detail to include. More so when you can not see the physical reaction at the other end. – Carl Christian Feb 07 '16 at 17:03
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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 $H_4-H_2<\ln 2$.

J.G.
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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

$$\frac{7}{12}<\log(2)<\frac{5}{6}$$

Community
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Jaume Oliver Lafont
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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

$$0<\frac{1}{2}<\frac{7}{12}<\frac{37}{60}<...\log(2)...<\frac{319}{420}<\frac{47}{60}<\frac{5}{6}<1$$

Jaume Oliver Lafont
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