Tomi's Challenges - #14

This exercise is all about logarithms! We will start by first simplifying $A$ .

$A=\begin{bmatrix} \sqrt[3]{\log_n(2^{\log_n2}n)} & 0\\ 0 & \sqrt[3]{\sum_{k=1}^{n} \frac{\log_n\frac{1}{2}}{k\cdot\ln2}} \end{bmatrix}^{\begin{bmatrix} 3 & 0\\ 0 & 3 \end{bmatrix}}=e^{{\begin{bmatrix} 3 & 0\\ 0 & 3 \end{bmatrix}\ln\begin{bmatrix} \sqrt[3]{\log_n(2^{\log_n2}n)} & 0\\ 0 & \sqrt[3]{\sum_{k=1}^{n} \frac{\log_n\frac{1}{2}}{k\cdot\ln2}} \end{bmatrix}}}=e^{{\begin{bmatrix} 3\ln\sqrt[3]{\log_n(2^{\log_n2}n)} & 0\\ 0 & 3\ln\sqrt[3]{\sum_{k=1}^{n} \frac{\log_n\frac{1}{2}}{k\cdot\ln2}} \end{bmatrix}}}=e^{{\begin{bmatrix} \ln\log_n(2^{\log_n2}n) & 0\\ 0 & \ln\sum_{k=1}^{n} \frac{\log_n\frac{1}{2}}{k\cdot\ln2} \end{bmatrix}}}={\begin{bmatrix} \log_n(2^{\log_n2}n) & 0\\ 0 & \sum_{k=1}^{n} \frac{\log_n\frac{1}{2}}{k\cdot\ln2} \end{bmatrix}}$ .

Next, we need to find the value we need to take the limit of. Since the trace is a linear operator, and $\frac{1}{log_n2}$ is a scalar from the field of the vector space of the matrix, we can conclude that $Tr(\frac{A}{log_n2})=\frac{1}{log_n2}Tr(A)=$

$\Large \frac{\log_n(2^{\log_n2}n)+\sum_{k=1}^{n} \frac{\log_n\frac{1}{2}}{k\cdot\ln2}}{log_n2}$

We need to take the limit of this value.

$\Large \lim_{n\to\infty} \frac{\log_n(2^{\log_n2}n)+\sum_{k=1}^{n} \frac{\log_n\frac{1}{2}}{k\cdot\ln2}}{log_n2}$

$\lim_{n\to\infty} \frac{1}{\log_n2}+\log_n2-\sum_{k=1}^{n} \frac{\log_n2}{k\cdot\ln2\log_n2}$

$\lim_{n\to\infty} \frac{1}{\log_2n}+\log_2n-\sum_{k=1}^{n} \frac{1}{k\cdot\ln2}$

If we assume the rest of the limit is convergent, we get

$\lim_{n\to\infty} \log_2n-\sum_{k=1}^{n} \frac{1}{k\cdot\ln2}$

$\lim_{n\to\infty} \frac{\ln n}{\ln2}-\sum_{k=1}^{n} \frac{1}{k\cdot\ln2}$

$\frac{1}{\ln2} \lim_{n\to\infty} \ln n-\sum_{k=1}^{n} \frac{1}{k}$

$-\frac{1}{\ln2} \lim_{n\to\infty} (\sum_{k=1}^{n} \frac{1}{k})-\ln n=-\frac{\gamma}{\ln2}\approx -0.8327$ , where $\gamma\approx 0.5772$ is the Euler-Mascheroni constant.