200 Followers Problem - A Logarithmic Integral!

First note that for $x \geq 1$ , we have:

$xa_k^{1/n} \leq (1+a_k x^n)^{1/n} \leq x(1+a_k)^{1/n} \quad ... (1)$

Since $a_k^{1/n}$ and $(1+a_k)^{1/n}$ each converge to $1$ as $n \to \infty$ , it follows from the above that $(1+a_kx^n)^{1/n}$ converges to $x$ as $n \to \infty$ . Thus,

$\lim_{n \to \infty} \dfrac{\ln(1+a_k x^n)}{n} = \lim_{n \to \infty} \ln(1+a_k x^n)^{1/n} = \ln(x) \quad ... (2)$

Taking logarithms across the last inequality in (1), we obtain;

$\dfrac{\ln(1+a_k x^n)}{n} \leq \ln(x) + \dfrac{\ln(1+a_k )}{n} \leq \ln(x) + \ln(1+a_k)$

from which it follows that:

$\prod_{k=1}^m \dfrac{\ln(1+a_k x^n)}{n} \leq \prod_{k=1}^m (\ln(x) + \ln(1+a_k))$

By Lebesgue's Dominated Convergence Theorem, we can bring the limits inside the integral ; then we apply (2) as follows:

$L_m = \int_1^e \lim_{n \to \infty} \prod_{k=1}^m \dfrac{\ln(1+a_k x^n)}{n} dx = \int_1^e \prod_{k=1}^m \lim_{n \to \infty} \dfrac{\ln(1+a_k x^n)}{n} dx = \int_1^e (\ln(x))^m dx \quad ...(3)$

Next, we'll integrate it by parts to obtain at the recurrence relation:

$\large{L_m = e - mL_{m-1}} \quad ...(4)$

Finally, we'll use induction on $m$ to show that (with the appropriate initial condition) the solution to the recurrence in (4) is given by;

$\large{L_m = (-1)^{m+1} m! + e \sum_{k=1}^m (-1)^k \dfrac{m!}{(m-k)!}}$

$\Longrightarrow \large{L_m = (-1)^{m+1} m! + (-1)^m eD_m}$

where $D_m$ is the number of derangements of $1,2, \ldots ,m$ .

Substituting $m=7$ , we obtain $L_7 = 5040 - 1854e$ .

Thus $A=5040, B= -1854, A+B = \boxed{3186}$ .

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Note that the answer doesn't depend on $a_k$ (s), so whatever it may be, even Fibonacci sequence, the answer still remains the same!

Relevant wiki: L'Hopital's Rule - Problem Solving

$\begin{aligned} L_m & = \lim_{n \to \infty} \frac 1{n^m} \int_1^e \prod_{k=1}^m \ln \left(1+a_kx^n\right) \ dx \\ & = \int_1^e \prod_{k=1}^m \left({\color{#3D99F6}\lim_{n \to \infty} \frac {\ln \left(1+a_kx^n \right)}n} \right) dx & \small \color{#3D99F6} \text{For }x > 1 \text{, a }\infty/\infty \text{ case, L'Hôpital's rule applies.} \\ & = \int_1^e \prod_{k=1}^m \left({\color{#3D99F6}\lim_{n \to \infty} \frac {a_k\ln x \cdot x^n}{1+a_kx^n}} \right) dx & \small \color{#3D99F6} \text{Differentiating up and down w.r.t }n \\ & = \int_1^e \prod_{k=1}^m \left({\color{#3D99F6}\lim_{n \to \infty} \frac {a_k\ln x}{\frac 1{x^n}+a_k}} \right) dx & \small \color{#3D99F6} \text{Dividing up and down by }x^n \\ & = \int_1^e \prod_{k=1}^{\color{#3D99F6}7} \ln x \ dx & \small \color{#3D99F6} \text{Putting }m=7 \\ & = \int_1^e \ln^7 x \ dx & \small \color{#3D99F6} \text{Let }u = \ln x \implies e^u \ du = dx \\ & = \int_0^1 u^7 e^u \ du & \small \color{#3D99F6} \text{By repeated integration by parts} \end{aligned}$

$\begin{aligned} \quad \ & = \left(u^7 - 7u^6+42u^5-210u^4+840u^3-2520u^2+5040u - 5040\right)e^u \bigg|_0^1 \\ & = 5040 - 1854e \end{aligned}$

$\implies A+B = 5040-1854 = \boxed{3186}$

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Generalization:

We note that $L_m$ is independent of $a_i$ and it is given by:

$\begin{aligned} L_m & = (-1)^{m+1} m! + e \sum_{k=0}^m (-1)^k \frac {m!}{(m-k)!} \end{aligned}$