Finding value of $\lim_{n\rightarrow \infty}\frac{1+\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n^3}}{\ln(n)}$ 
Find the value of
  $$\lim_{n\rightarrow \infty}\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots  +\frac{1}{n^3}}{\ln(n)}$$

My Try: Using Stolz-Cesaro,
Let $\displaystyle a_{n} = 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots +\frac{1}{n}$
 and $b_{n} = \ln(n)$
So $\displaystyle \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} = \lim_{n\rightarrow \infty}\frac{1}{{(n+1)^3}}\cdot \frac{1}{\ln\bigg(1+\frac{1}{n}\bigg)} = 0$
Please explain if what I have done above is right.

To answer your question directly. The mistake is made in $a_{n+1}-a_n$. For example, if $n = 10$, then this is given by
$$a_{11}-a_{10} = 1/1331 + 1/1330 + \cdots + 1/1001$$
since $11^3 = 1331$ and $10^3=1000$.
If you insert this, you still are going to need to do something similar as Gabriel suggested by viewing the summation as upper and lower Riemann sums.
Edit2: if you now use Stolz Cesaro, which is possible, it works out as Gimusi did.
