Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

Approach :

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

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


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, $0.3 < 1$

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


share|cite|improve this question
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
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.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.