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Prove the statement : $\log(k + 1) - \log k > \frac{3}{10k}$

Approach :

$$\log(k+1)-\log{k} > \frac{3}{10k}$$

Clearly, $k\in\mathbb{Z}^{+}$

$$\log(k+1)-\log{k}=\log\bigg(1+\frac{1}{k}\bigg)$$

given base is $10$, so

$$\log\left(1+\frac{1}{k}\right) > \log\left(\frac{1}{k}\right) \implies \log\left(1+\frac{1}{k}\right) > \frac{1}{k}$$

Since, $0.3 < 1$

$$ \log\left(1+\frac{1}{k}\right) > \frac{3}{10k}$$

QED

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I'm not satisfied with this proof. Please help. –  Hyperbola Jul 12 '12 at 10:25
    
And it is a good thing that you are not since, in fact, $\log(1+1/k)\lt1/k$ for every positive $k$. –  Did Jul 14 '12 at 14:13

2 Answers 2

up vote 2 down vote accepted

Solution №1.

Consider function $f(x)=\log(x)$ and fix $k\in\mathbb{Z}_+$. By mean value theorem there exist $c\in[k,k+1]$ such that $$ \log(k+1)-\log(k)=(\log x)'|_{x=c}((k+1)-k)=\frac{1}{c} $$ Since $c>k+1$ then $$ \log(k+1)-\log(k)=\frac{1}{c}>\frac{1}{k+1} $$ Since $k\in\mathbb{Z}_+$, then $k+1<10/3k$ and we obtain $$ \log(k+1)-\log(k)>\frac{1}{k+1}>\frac{3}{10 k} $$

Solution №2.

It is enough to show that $\log(1+x)>0.3x$ for all $x\in (0,1)$. Then you can take $x=1/k$ for each $k\in\mathbb{Z}_+$ and prove your inequality.

In order to prove inequality $\log(1+x)>0.3x$ for all $x\in (0,1)$, consider function $$ f(x)=\log(1+x)-0.3x $$ You can check, that

  • $f(0)=0$
  • $f'(x)=\frac{0.7-0.3x}{x+1}>0$ for $x\in (0,1)$.

Hence $f$ is non-negative on $(0,1)$, which is equivalent to $$ \log(1+x)>0.3x\quad\text{ for }\quad x\in(0,1) $$ The rest is clear.

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$$\log(k+1)-\log(k)=\int_k^{k+1}\frac{\mathrm dx}x\gt\int_k^{k+1}\frac{\mathrm dx}{k+1}=\frac1{k+1}\geqslant\frac1{2k}\qquad(k\geqslant1)$$ ...and, likewise, $$\log(k+1)-\log(k)\lt\int_k^{k+1}\frac{\mathrm dx}{k}=\frac1{k}\qquad(k\geqslant1)$$

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