How to prove that $\lim_{n\to\infty}\int_k^n\frac{\log_xn}{n\log x}\, dx=0$ for every $k\ge 2$

Question

Prove that $$\lim_{n\to\infty}\int_k^n\frac{\log_xn}{n\log x}\, dx=0$$ for every $k\ge 2$.

The integral can be written as $$\int_k^n\frac{\log n}{n\log^2x}\, dx=\frac{\log n}n\int_k^n\frac1{\log^2x}\, dx$$ Now since $e^n>n$ for $n>1$, $\dfrac{\log n}n\to 0$ as $n\to\infty$.

Also, $\dfrac1{\log^2x}$ is monotonically decreasing, so we need only consider the case when $k=2$ - the equality would hold for larger values of $k$.

However, by WolframAlpha (http://www.wolframalpha.com/input/?i=%5Cint_2%5E%5Cinfty%5Cfrac1%7B%5Clog%5E2x%7D+dx), $$\int_2^n\frac1{\log^2x}\, dx\to +\infty$$ as $n\to\infty$. The limit thus cannot be evaluated using this method because we would have $0\cdot\infty$.

How should I begin the proof?

Answer 1

Write your expression in the form $$\frac{\displaystyle\int_k^n\dfrac{dx}{\log^2(x)}}{\dfrac n{\log(n)}}$$ Since this is defined for real $n>1$ you can apply L'Hôpital's rule to obtain: $$\frac{\dfrac 1{\log^2(n)}}{\dfrac{\log(n)-1}{\log^2(n)}}=\frac 1{\log(n)-1}\sim\frac 1{\log(n)}\longrightarrow 0$$ as $n\to +\infty$.