[math]Tokyo Institute of Technology, Entrance exam, math, 2019

math

Photo by Vlad Bagacian on Pexels.com

2023.01.03

Table of Contents

Problem

$\displaystyle a = \frac{2^8}{3^4}$，consider the number sequence $$b_k = \frac{(k+1)^{k+1}}{a^k k!}\ \ (k = 1, 2, 3, \cdots)$$ as Show that the
$(1)$ function $\displaystyle f(x) = (x+1)\log{\left(1+\frac{1}{x}\right)}$ decreases with $x > 0$.
$(2)$ Express the maximum value of the term $M$ of the number sequence $\{b_k\}$ as an irreducible fraction and find all $k$ such that $b_k = M$.

Plan

We’ll let ourselves be guided.

Answer

$(1)$$$\begin{eqnarray}f(x) & = & (x+1)\log{\left(1+\frac{1}{x}\right)}\\ & = & (x+1)\log\left(\frac{x+1}{x}\right)\\ & = & (x+1)\log{(x+1 )}-(x+1)\log{x}\end{eqnarray}$$, so $$\begin{eqnarray}f^{\prime}(x) & = & \log{(x+1)}+(x+1)\cdot \frac{1}{x+1}-\log{x}-(x+1)\cdot \frac{1}{x}\\ & = & \log{(x+1)}-\log{x}-\frac{1}{x}\\ & = & \log{\left(1+\frac{1}{x}\right)}-\frac{1}{x}\end{eqnarray}$$ In general, since $\log{(1+t)}\leq t$ for positive numbers $t$ (the equality holds for $t = 0$), we know that$f^{\prime}(x) < 0$. In other words, we know that $f(x)$ is a decreasing function.

$(2)$$$\begin{eqnarray}\frac{b_{k+1}}{b_k} & = & \frac{(k+2)^{k+2}}{a^{k+1}(k+1)!} \cdot \frac{a^k k!}{(k+1)^{k+1}}\\ & = & \frac{k+2}{a(k+1)}\cdot \left(1+\frac{1}{k+1}\right)^{k+1}\\ & = & \frac{1}{a}\cdot \left(1+\frac{1}{a k+1}\right)^{k+2}\end{eqnarray}$$ We put this as $c_k$. Since $\displaystyle f(x) = \log{\left(1+\frac{1}{x}\right)^{x+1}}$ is a decreasing function, $\displaystyle \left(1+\frac{1}{x}\right)^{x+1}$ is also is also a decreasing function. Therefore, $c_k$ is a decreasing sequence. $$\begin{eqnarray}c_1 & = & \frac{3^4}{2^8}\cdot \left(1+\frac{1}{2}\right)^3\\ & = & \frac{3^7}{2^{11}}\\ & = & \frac{2187}{2048}\\ & > & 1\\ c_2 & = & \frac{3^4}{2^8}\cdot \left(1+\frac{1}{3}\right)^4\\ & = & 1\\ c_3 &= & \frac{3^4}{2^8}\cdot \left(1+\frac{1}{4}\right)^5\\ & = & \frac{3^45^5}{2^{18}}\\ & = & \frac{253125}{262144}\\ & < & 1\end{eqnarray}$$, so $$\begin{eqnarray}\frac{b_2}{b_1} & > & 1\\ \frac{b_3}{b_2} & = & 1\\ \frac{b_4}{b_3} & < & 1\end{eqnarray}$$, and since $c_k < 1$ when $k\geq 3$, we have $b_3 > b_4 > \cdots$. Since $$\begin{eqnarray}b_2 & = & \frac{(2+1)^{2+1}}{a^22!}\\ & = & \frac{3^3\cdot 3^8}{2^{16}\cdot 2}\\ & = & \frac{177147}{131072}\end{eqnarray}$$, then Therefore, the maximum value of $b_k$ is $\displaystyle \underline{\frac{177147}{131072}}$ and the maximum value of $b_k = M$ is $\underline{k = 2, 3}$.