Prove the statement : $\log(k + 1) -\log k>\frac{ 3}{10k}$ 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
 A: 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.
A: $$\log(k+1)-\log(k)=\int_k^{k+1}\frac{\mathrm dx}x\gt\int_k^{k+1}\frac{\mathrm dx}{k+1}=\frac1{k+1}\geqslant\frac1{2k}\qquad(k\geqslant1)$$
...and, likewise,
$$\log(k+1)-\log(k)\lt\int_k^{k+1}\frac{\mathrm dx}{k}=\frac1{k}\qquad(k\geqslant1)$$
