Question

Let $f(x) = \lim_{n\to\infty} \frac{n^n(x+n)(x+\frac{n}{2})...(x+\frac{n}{n})}{n!(x^2+n^2)(x^2+\frac{n^2}{4})...(x^2+\frac{n^2}{n^2})}$, for all $x>0$. Then

• A. $f(\frac{1}{2})\ge f(1)$
• B. $f(\frac{1}{3})\le f(\frac{2}{3})$
• C. $f'(2)\le 0$
• D. $\frac{f'(3)}{f(3)}\ge \frac{f'(2)}{f(2)}$

Answer 1

Answer: (A), (D)

A, D

Answer 2

To determine $f(x)$, we simplify the given expression: $f(x) = \lim_{n\to\infty} \frac{n^n(x+n)(x+\frac{n}{2})...(x+\frac{n}{n})}{n!(x^2+n^2)(x^2+\frac{n^2}{4})...(x^2+\frac{n^2}{n^2})}$ The numerator can be written as $N = n^n \prod_{k=1}^{n} (x+\frac{n}{k})$. The denominator can be written as $D = n! \prod_{k=1}^{n} (x^2+\frac{n^2}{k^2})$.

Let's factor out $n$ from each term in the numerator product and $n^2$ from each term in the denominator product: $\prod_{k=1}^{n} (x+\frac{n}{k}) = \prod_{k=1}^{n} n(\frac{x}{n}+\frac{1}{k}) = n^n \prod_{k=1}^{n} (\frac{x}{n}+\frac{1}{k})$. $\prod_{k=1}^{n} (x^2+\frac{n^2}{k^2}) = \prod_{k=1}^{n} n^2(\frac{x^2}{n^2}+\frac{1}{k^2}) = n^{2n} \prod_{k=1}^{n} (\frac{x^2}{n^2}+\frac{1}{k^2})$.

Substitute these back into the expression for $f(x)$: $f(x) = \lim_{n\to\infty} \frac{n^n \cdot n^n \prod_{k=1}^{n} (\frac{x}{n}+\frac{1}{k})}{n! \cdot n^{2n} \prod_{k=1}^{n} (\frac{x^2}{n^2}+\frac{1}{k^2})} = \lim_{n\to\infty} \frac{1}{n!} \frac{n^{2n}}{n^{2n}} \prod_{k=1}^{n} \frac{\frac{x}{n}+\frac{1}{k}}{\frac{x^2}{n^2}+\frac{1}{k^2}}$ $f(x) = \lim_{n\to\infty} \frac{1}{n!} \prod_{k=1}^{n} \frac{k(x+n/k)}{k^2(x^2+n^2/k^2)} \cdot k$ This simplification was incorrect in my scratchpad. Let's restart the main simplification part.

Let's use the form $f(x) = \lim_{n\to\infty} \frac{\prod_{k=1}^{n} (1+\frac{kx}{n})}{\prod_{k=1}^{n} (1+\frac{k^2x^2}{n^2})}$. This was derived correctly. Let $P_1 = \prod_{k=1}^{n} (1+\frac{kx}{n})$ and $P_2 = \prod_{k=1}^{n} (1+\frac{k^2x^2}{n^2})$. We evaluate $\lim_{n\to\infty} \ln P_1$ and $\lim_{n\to\infty} \ln P_2$ using Riemann sums. $\lim_{n\to\infty} \ln P_1 = \lim_{n\to\infty} \sum_{k=1}^{n} \ln(1+\frac{kx}{n}) = \int_0^1 \ln(1+tx) dt$. Let $u=1+tx$, $du=x dt$. When $t=0, u=1$. When $t=1, u=1+x$. $\int_0^1 \ln(1+tx) dt = \frac{1}{x} \int_1^{1+x} \ln u du = \frac{1}{x} [u\ln u - u]_1^{1+x}$ $= \frac{1}{x} [(1+x)\ln(1+x) - (1+x) - (1\ln 1 - 1)] = \frac{1}{x} [(1+x)\ln(1+x) - x]$. So, $\lim_{n\to\infty} P_1 = e^{\frac{(1+x)\ln(1+x)-x}{x}} = e^{\ln((1+x)^{(1+x)/x})} e^{-1} = \frac{(1+x)^{(1+x)/x}}{e}$.

$\lim_{n\to\infty} \ln P_2 = \lim_{n\to\infty} \sum_{k=1}^{n} \ln(1+\frac{k^2x^2}{n^2}) = \int_0^1 \ln(1+t^2x^2) dt$. Let $u=tx$, $du=x dt$. When $t=0, u=0$. When $t=1, u=x$. $\int_0^1 \ln(1+t^2x^2) dt = \frac{1}{x} \int_0^x \ln(1+u^2) du$. Using integration by parts: $\int \ln(1+u^2) du = u\ln(1+u^2) - \int u \frac{2u}{1+u^2} du = u\ln(1+u^2) - 2\int \frac{u^2}{1+u^2} du$ $= u\ln(1+u^2) - 2\int (1 - \frac{1}{1+u^2}) du = u\ln(1+u^2) - 2u + 2\arctan u$. So, $\frac{1}{x} [u\ln(1+u^2) - 2u + 2\arctan u]_0^x = \frac{1}{x} [x\ln(1+x^2) - 2x + 2\arctan x]$. So, $\lim_{n\to\infty} P_2 = e^{\frac{x\ln(1+x^2)-2x+2\arctan x}{x}} = e^{\ln(1+x^2)} e^{-2} e^{\frac{2\arctan x}{x}} = (1+x^2) e^{-2} e^{\frac{2\arctan x}{x}}$.

Therefore, $f(x) = \frac{\frac{(1+x)^{(1+x)/x}}{e}}{(1+x^2) e^{-2} e^{\frac{2\arctan x}{x}}} = \frac{(1+x)^{(1+x)/x}}{1+x^2} e^{1 - \frac{2\arctan x}{x}}$.

Let's analyze $f(x)$. Let $g(x) = \ln f(x) = \frac{1+x}{x}\ln(1+x) - \ln(1+x^2) + 1 - \frac{2\arctan x}{x}$. $g(x) = (1+\frac{1}{x})\ln(1+x) - \ln(1+x^2) + 1 - \frac{2\arctan x}{x}$.

To check the options, we need $f'(x)$. It's easier to find $g'(x) = \frac{f'(x)}{f(x)}$. $g'(x) = \left(-\frac{1}{x^2}\ln(1+x) + (1+\frac{1}{x})\frac{1}{1+x}\right) - \frac{2x}{1+x^2} - 2\left(-\frac{1}{x^2}\arctan x + \frac{1}{x}\frac{1}{1+x^2}\right)$ $g'(x) = -\frac{\ln(1+x)}{x^2} + \frac{1}{x} - \frac{2x}{1+x^2} + \frac{2\arctan x}{x^2} - \frac{2}{x(1+x^2)}$ $g'(x) = \frac{2\arctan x - \ln(1+x)}{x^2} + \frac{1}{x} - \frac{2x}{1+x^2} - \frac{2}{x(1+x^2)}$ $g'(x) = \frac{2\arctan x - \ln(1+x)}{x^2} + \frac{x(1+x^2) - 2x^2 - 2x}{x^2(1+x^2)}$ $g'(x) = \frac{2\arctan x - \ln(1+x)}{x^2} + \frac{x+x^3 - 2x^2 - 2x}{x^2(1+x^2)}$ $g'(x) = \frac{2\arctan x - \ln(1+x)}{x^2} + \frac{x^3 - 2x^2 - x}{x^2(1+x^2)}$ $g'(x) = \frac{2\arctan x - \ln(1+x)}{x^2} + \frac{x^2 - 2x - 1}{x(1+x^2)}$.

Let's recheck the derivative of $\frac{1+x}{x}\ln(1+x)$. $\frac{d}{dx} \left( (\frac{1}{x}+1)\ln(1+x) \right) = (-\frac{1}{x^2})\ln(1+x) + (\frac{1}{x}+1)\frac{1}{1+x} = -\frac{\ln(1+x)}{x^2} + \frac{1}{x}$. This is correct. Let's recheck the derivative of $\frac{2\arctan x}{x}$. $\frac{d}{dx} \left( \frac{2\arctan x}{x} \right) = 2 \frac{\frac{1}{1+x^2}x - \arctan x}{x^2} = \frac{2x}{(1+x^2)x^2} - \frac{2\arctan x}{x^2} = \frac{2}{x(1+x^2)} - \frac{2\arctan x}{x^2}$. So, $-\frac{d}{dx} \left( \frac{2\arctan x}{x} \right) = -\frac{2}{x(1+x^2)} + \frac{2\arctan x}{x^2}$. This is correct.

So, $g'(x) = -\frac{\ln(1+x)}{x^2} + \frac{1}{x} - \frac{2x}{1+x^2} - \left(\frac{2}{x(1+x^2)} - \frac{2\arctan x}{x^2}\right)$ $g'(x) = \frac{2\arctan x - \ln(1+x)}{x^2} + \frac{1}{x} - \frac{2x}{1+x^2} - \frac{2}{x(1+x^2)}$. This is correct. $g'(x) = \frac{2\arctan x - \ln(1+x)}{x^2} + \frac{x(1+x^2) - 2x^2 - 2}{x(1+x^2)}$. $g'(x) = \frac{2\arctan x - \ln(1+x)}{x^2} + \frac{x^3 - 2x^2 + x - 2}{x(1+x^2)}$. $g'(x) = \frac{2\arctan x - \ln(1+x)}{x^2} + \frac{x^2(x-2) + (x-2)}{x(1+x^2)} = \frac{2\arctan x - \ln(1+x)}{x^2} + \frac{(x^2+1)(x-2)}{x(1+x^2)}$ $g'(x) = \frac{2\arctan x - \ln(1+x)}{x^2} + \frac{x-2}{x} = \frac{2\arctan x - \ln(1+x) + x(x-2)}{x^2}$. $g'(x) = \frac{2\arctan x - \ln(1+x) + x^2 - 2x}{x^2}$.

Let $h(x) = 2\arctan x - \ln(1+x) + x^2 - 2x$. We need to analyze the sign of $h(x)$. $h'(x) = \frac{2}{1+x^2} - \frac{1}{1+x} + 2x - 2$. $h'(x) = \frac{2(1+x) - (1+x^2)}{(1+x^2)(1+x)} + 2(x-1) = \frac{2+2x-1-x^2}{(1+x^2)(1+x)} + 2(x-1) = \frac{1+2x-x^2}{(1+x^2)(1+x)} + 2(x-1)$. For $x>0$: $h(1) = 2\arctan 1 - \ln 2 + 1 - 2 = 2(\pi/4) - \ln 2 - 1 = \pi/2 - \ln 2 - 1 \approx 1.57 - 0.693 - 1 = -0.123 < 0$. $h(2) = 2\arctan 2 - \ln 3 + 4 - 4 = 2\arctan 2 - \ln 3 \approx 2(1.107) - 1.098 = 2.214 - 1.098 = 1.116 > 0$. Since $h(1)<0$ and $h(2)>0$, there is a root between 1 and 2.

Consider $x=1$: $g'(1) = \frac{h(1)}{1^2} = \pi/2 - \ln 2 - 1 < 0$. Since $f'(x) = f(x) g'(x)$ and $f(x)>0$ for $x>0$, the sign of $f'(x)$ is the same as $g'(x)$. So $f'(1) < 0$.

Let's check the options. (A) $f(\frac{1}{2})\ge f(1)$. This means $f(x)$ is decreasing from $x=1/2$ to $x=1$. (B) $f(\frac{1}{3})\le f(\frac{2}{3})$. This means $f(x)$ is increasing from $x=1/3$ to $x=2/3$. (C) $f'(2)\le 0$. This means $g'(2) \le 0$. $h(2) = 2\arctan 2 - \ln 3 \approx 1.116 > 0$. So $g'(2) > 0$. Thus $f'(2) > 0$. Option (C) is false.

(D) $\frac{f'(3)}{f(3)}\ge \frac{f'(2)}{f(2)}$. This is $g'(3) \ge g'(2)$. We know $g'(2) = \frac{h(2)}{4} = \frac{2\arctan 2 - \ln 3}{4} \approx \frac{1.116}{4} = 0.279$. $h(3) = 2\arctan 3 - \ln 4 + 9 - 6 = 2\arctan 3 - \ln 4 + 3$. $2\arctan 3 \approx 2(1.249) = 2.498$. $\ln 4 = 2\ln 2 \approx 2(0.693) = 1.386$. $h(3) \approx 2.498 - 1.386 + 3 = 4.112 > 0$. $g'(3) = \frac{h(3)}{9} \approx \frac{4.112}{9} \approx 0.457$. Since $0.457 \ge 0.279$, option (D) is true.

Let's confirm the monotonicity of $h(x)$. $h'(x) = \frac{1+2x-x^2}{(1+x^2)(1+x)} + 2(x-1)$. For $x>1$, $x-1>0$. The term $\frac{1+2x-x^2}{(1+x^2)(1+x)}$: The numerator $1+2x-x^2$ is a downward parabola with roots $x = \frac{-2 \pm \sqrt{4-4(-1)(1)}}{-2} = \frac{-2 \pm \sqrt{8}}{-2} = 1 \mp \sqrt{2}$. So $1+2x-x^2 > 0$ for $1-\sqrt{2} < x < 1+\sqrt{2}$. $1+\sqrt{2} \approx 1+1.414 = 2.414$. So for $x \in (1, 1+\sqrt{2})$, $h'(x) > 0$. Since $h(1)<0$ and $h(2)>0$, and $h'(x)>0$ for $x \in (1, 2.414)$, it means $h(x)$ is increasing in this interval. So $g'(x)$ increases from $x=1$ to $x \approx 2.414$. Since $g'(2)>0$ and $g'(3)>0$, and $g'(x)$ is increasing in $(1, 2.414)$, and $g'(x)$ decreases after $2.414$, we need to compare $g'(2)$ and $g'(3)$. The root of $h(x)$ is between 1 and 2. Let's call it $x_0$. For $x>x_0$, $h(x)>0$, so $g'(x)>0$. For $x<x_0$, $h(x)<0$, so $g'(x)<0$. This means $f(x)$ is decreasing for $x \in (0, x_0)$ and increasing for $x \in (x_0, \infty)$. Since $x_0 \in (1,2)$, $f(1/2) > f(1)$ because $1/2 < 1 < x_0$. This makes option (A) true. Also $f(1/3) < f(2/3)$ is true if $1/3$ and $2/3$ are both less than $x_0$. Since $x_0 > 1$, $1/3 < 2/3 < 1 < x_0$. So $f(x)$ is decreasing in this interval. Therefore $f(1/3) > f(2/3)$. So option (B) is false.

Let's recheck option (A) and (B). $f(x)$ is decreasing for $x \in (0, x_0)$ where $x_0 \in (1,2)$. (A) $f(1/2) \ge f(1)$. Since $1/2 < 1$ and both are in $(0, x_0)$, $f(x)$ is decreasing. So $f(1/2) > f(1)$. Option (A) is true. (B) $f(1/3) \le f(2/3)$. Since $1/3 < 2/3$ and both are in $(0, x_0)$, $f(x)$ is decreasing. So $f(1/3) > f(2/3)$. Option (B) is false.

So (A) and (D) are correct.