1
$\begingroup$

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

  • 0
    And it is a good thing that you are not since, in fact, $\log(1+1/k)\lt1/k$ for every positive $k$.2012-07-14

2 Answers 2

2

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.

2

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