<s>[INST] 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.
 [/INST] Your $a_n$ should go up to $\frac1{n^3}$. Then $a_{n+1}-a_n=\frac{1}{(n+1)^3}+\frac{1}{(n+1)^3-1}\dots+\frac1{n^3+2}+\frac{1}{n^3+1}$. 
I don't know whether you can arrive there by your method, but you can find the limit by using the asymptotics of harmonic numbers / definition of the euler $\gamma$ constant. Then the limit should be $3$.
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