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.