$\limsup_{k \to \infty}\frac{u_k}{\ln(k)}=1 \implies \lim_{k \to \infty}\frac{1}{\ln(k)}\max_{0 \leq r \leq k}u_r=1$

Question

Let $(u_k)_{k \in \mathbb{N}}$ be a sequence of $\mathbb{R}_+$ such that $\limsup_{k \to \infty}\frac{u_k}{\ln(k)}=1.$

Prove or disprove the following: $\lim_{k \to \infty}\frac{1}{\ln(k)}\max_{0 \leq r \leq k}u_r=1.$

Any suggestions how to begin?

Answer 1

Consider $m_n = [e^{e^n}]$, $u_k = \sum_{n \ge 1} I_{ k = m_n} \cdot \ln m_n$, where $I$ is an indicator function. So almost all $u_k$ are equal to $0$, but if $k = [e^{e^1}], [e^{e^2}], [e^{e^3}], \ldots $ then $u_k > 0$ and if $k = [e^{e^n}]$ then $u_k = \ln k$.

Hence $\overline{\lim}_k \frac{u_k}{\ln k} = 1$. Put $b_k = \frac{\max_{0 \le r \le k} u_k}{\ln k}$, then $b_k \not \to 1$ because $$b_{(m_n - 1)} = \frac{\max_{0 \le k \le (m_n - 1)} u_k}{\ln(m_n - 1)} = \frac{\max_{0 \le k \le m_{n-1}} u_k}{\ln(m_n - 1)} = $$ $$=\frac{u_{m_{n - 1}}}{\ln(m_n - 1)} = \frac{\ln m_{n - 1}}{\ln(m_n - 1)} = \frac{ln[e^{e^{n-1}}]}{ln[e^{e^{n}}-1]} \to e^{-1}$$ because $ln[e^{e^{n-1}}] = ln ( e^{e^{n-1}} ) + o(1) = e^{n-1} + o(1)$ $\big($as $ln \frac{( e^{e^{n-1}} )}{[e^{e^{n-1}}]} = \ln (1+o(1))$ $\big)$ and $ln[e^{e^{n}}-1] = ln (e^{e^{n}}) + o(1) = e^{n} + o(1) $.