Somemathhere

$\textbf{Problem.}$  Let $f:[0,1]\to\mathbb{R}$ be a continuous function. Find the limit  $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right).$$   $\textbf{Solution.}$  Fix $\varepsilon>0$. We aim to show that for large enough $n$   $$\left|\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|<3\varepsilon$$  which would mean that the limit is $f(1).$ Let $M$ be an upper bound for $|f|$ on $[0,1]$, i.e. $|f(x)|\le M$ for all $x\in [0,1].$  For all $n\ge 1$ define $$k(n)=\left\lfloor\frac{n}{n^{1/\sqrt{\ln n}}}\right\rfloor.$$ The number is chosen in such a way that $$\lim_{n\to\infty}\frac{k(n)}{n}=0 \ \ \& \ \  \lim_{n\to\infty}\frac{\ln(k(n))}{\ln n}=1.$$  Split the sum in two: $$\left|\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|\le \left|\frac{1}{n}\sum_{k=1}^{k(n)} f\left(\frac{\ln k}{\ln n }\right)\right|+\left|\frac{1}{n}\sum_{k=k(n)+1}^{n} f\left(\frac{\ln k}{\ln n }\right)-f(1)\right| \ \ \ (*)$$  T