Evaluate $\lim\limits_{n \to \infty}\frac{\frac{\ln2}{2}+\frac{\ln3}{3}+\cdots+\frac{\ln n}{n}}{\ln^2 n}$.

Question

Problem

Evaluate $$\lim_{n \to \infty}\dfrac{\dfrac{\ln2}{2}+\dfrac{\ln3}{3}+\cdots+\dfrac{\ln n}{n}}{\ln^2 n}.$$

Thanks for @LittleCuteKemono's hint. Here is my solution. Please correct me if I'm wrong!

According to Stoltz-Cesaro Theorem, we have $$\lim_{n \to \infty}\dfrac{\dfrac{\ln2}{2}+\dfrac{\ln3}{3}+\cdots+\dfrac{\ln n}{n}}{\ln^2 n}=\lim_{n \to \infty}\dfrac{\dfrac{\ln (n+1)}{n+1}}{\ln^2 (n+1)-\ln^2 n}.$$Consider the function $f(x)=\ln^2 x.$ Notice that $f'(x)=2\cdot \dfrac{\ln x}{x}.$ By Lagrange's Mean Value Theorem, we have $$\ln^2(n+1)-\ln^2 n=f'(\xi)(n+1-n)=f'(\xi)=2\cdot \frac{\ln \xi}{\xi},$$where $n<\xi<n+1.$ Morover, consider another function $g(x)=\dfrac{\ln x}{x}.$ Since $g'(x)=\dfrac{1-\ln x}{x^2}<0$ holds for all $x>e,$ hence $g(n+1)<g(\xi)<g(n)$ holds for every sufficiently large $n.$ Therefore, $$\frac{1}{2} \leftarrow\frac{1}{2}\cdot\dfrac{g(n+1)}{g(n)}<\dfrac{\dfrac{\ln (n+1)}{n+1}}{\ln^2 (n+1)-\ln^2 n}=\frac{1}{2}\cdot\dfrac{g(n+1)}{g(\xi)}<\frac{1}{2}\cdot\dfrac{g(n+1)}{g(n+1)}=\frac{1}{2}.$$ Thus, by Squeeze Theorem, we have that the limit we want euqals $\dfrac{1}{2}.$

Answer 1

Note that $\frac d{dx}\ln^2x=\frac{2\ln x}{x}$ and that $\frac{\ln x}x$ is decreasing for $x>e$ (because $\frac d{dx}\frac{\ln x}x=\frac{1-\ln x}{x^2}$). Therefore, $$\left.\frac12\ln^2x\right|_4^{n+1}=\int_4^{n+1}\frac{\ln x}{x}\,\mathrm dx<\frac{\ln4}4+\ldots+\frac{\ln n}n< \int_3^{n}\frac{\ln x}{x}\,\mathrm dx=\left.\frac12\ln^2x\right|_3^{n}$$

Answer 2

Thanks for @LittleCuteKemono's hint. Here is my solution. Am I right?

According to Stoltz-Cesaro Theorem, we have $$\lim_{n \to \infty}\dfrac{\dfrac{\ln2}{2}+\dfrac{\ln3}{3}+\cdots+\dfrac{\ln n}{n}}{\ln^2 n}=\lim_{n \to \infty}\dfrac{\dfrac{\ln (n+1)}{n+1}}{\ln^2 (n+1)-\ln^2 n}.$$Consider the function $f(x)=\ln^2 x.$ Notice that $f'(x)=2\cdot \dfrac{\ln x}{x}.$ By Lagrange's Mean Value Theorem, we have $$\ln^2(n+1)-\ln^2 n=f'(\xi)(n+1-n)=f'(\xi)=2\cdot \frac{\ln \xi}{\xi},$$where $n<\xi<n+1.$ Morover, consider another function $g(x)=\dfrac{\ln x}{x}.$ Since $g'(x)=\dfrac{1-\ln x}{x^2}<0$ holds for all $x>e,$ hence $g(n+1)<g(\xi)<g(n)$ holds for every sufficiently large $n.$ Therefore, $$\frac{1}{2} \leftarrow\frac{1}{2}\cdot\dfrac{g(n+1)}{g(n)}<\dfrac{\dfrac{\ln (n+1)}{n+1}}{\ln^2 (n+1)-\ln^2 n}=\frac{1}{2}\cdot\dfrac{g(n+1)}{g(\xi)}<\frac{1}{2}\cdot\dfrac{g(n+1)}{g(n+1)}=\frac{1}{2}.$$ Thus, by Squeeze Theorem, we have that the limit we want euqals $\dfrac{1}{2}.$