# Solving $\lim_\limits{x\to\infty}{f(x)}=x^a{\left(1-\frac {0.1\prod_\limits{t=1}^x{0.5}^{1/t}}{10}\right)}^{x-a}$

I have asked about a limit before, which is still unanswered. With few modifications we can arrive to the following limit, which experimentally can be be shown to converge to $0$. However, I can't prove it.

$$\lim_\limits{x\to\infty}{f(x)}=x^a{\left(1-\frac {0.1\prod_\limits{t=1}^x{0.5}^{1/t}}{10}\right)}^{x-a},$$

where $a\in\mathbb Z^+$ is constant and $x\in\mathbb Z^+$.

I think we can solve this limit by solving the following one:

$$\lim_\limits{x\to\infty}{f(x)}=x^a{\left(1-\frac {0.1\cdot {0.5}^{\ln k}}{10}\right)}^{x-a}.$$

For large $x$ we know that $H(x)=\sum_{t=1}^{x}\frac{1}{t}\approx \ln(x)$. Your limit will be equal to $$x^a\left(1-\frac{1}{100\cdot x^{\ln(2)}}\right)^{x-a}=x^a e^{(x-a)\ln\left(1-\frac{1}{100\cdot x^{\ln(2)}}\right)}$$ Since $\ln(1+y)=y+O(y^2)$ when $y\to 0$, we have:

$$(x-a)\ln\left(1-\frac{1}{100\cdot x^{\ln(2)}}\right)=-\frac{(x-a)}{100\cdot x^{\ln(2)}}+O\left(x^{1-2\ln(2)}\right)$$ when $x\to \infty$, so we have $$\lim_{x\to\infty} f(x)=\lim_{x\to\infty}e^{-\frac{a(x-a)\ln(x)}{100\cdot x^{\ln(2)}}+O\left(x^{1-2\ln(2)}\ln(x)\right)}$$ Obviously that when $x\to\infty$ exponent's power $-\frac{a(x-a)\ln(x)}{100\cdot x^{\ln(2)}}+O\left(x^{1-2\ln(2)}\ln(x)\right)\to -\infty$ and our limit will be zero

Write $f(x)=x^a{\left(1-\frac {.1\prod_\limits{t=1}^x{0.5}^{1/t}}{10}\right)}^{x-a}$ as $f(x)=x^a{\left(1-c\prod_\limits{t=1}^xd^{1/t}\right)}^{x-a}$ where $c = 1/100$ and $d = 0.5$.

Since $\prod_\limits{t=1}^xd^{1/t} =\exp(\sum_\limits{t=1}^x\frac{\ln(d)}{t}) =\exp(\ln(d)\sum_\limits{t=1}^x\frac{1}{t}) =\exp(\ln(d)H_x) =d^{H_x}$ where, for large $x$, $H_x = \sum_\limits{t=1}^x\frac{1}{t} =\ln(x)+\gamma+O(1/x)$ (the well-known harmonic sum),

$\begin{array}\\ f(x) &=x^a{\left(1-cd^{H_x}\right)}^{x-a}\\ &=x^a{\left(1-cd^{\ln(x)+\gamma+O(1/x)}\right)}^{x-a}\\ &=x^a{\left(1-cd^{\ln(x)}d^{\gamma}d^{O(1/x)}\right)}^{x-a}\\ &=x^a{\left(1-cx^{\ln(d)}d^{\gamma}d^{O(1/x)}\right)}^{x-a}\\ \end{array}$

so, for large $x$,

$\begin{array}\\ \ln(f(x)) &=a\ln(x)+(x-a)\ln(1-cx^{\ln(d)}d^{\gamma}d^{O(1/x)})\\ &=a\ln(x)+(x-a)\ln(1-cx^{\ln(d)}d^{\gamma}(1+O(1/x)))\\ &=a\ln(x)+(x-a)\ln(1-cx^{\ln(d)}d^{\gamma}+O(x^{\ln(d)-1}))\\ &=a\ln(x)-(x-a)(cx^{\ln(d)}d^{\gamma}+O(x^{\ln(d)-1})+O(x^{2\ln(d)}))\\ &=a\ln(x) -x(cx^{\ln(d)}d^{\gamma}+O(x^{\ln(d)-1})+O(x^{2\ln(d)})) +a(cx^{\ln(d)}d^{\gamma}+O(x^{\ln(d)-1})+O(x^{2\ln(d)}))\\ &=a\ln(x) -cx^{1+\ln(d)}d^{\gamma}+O(x^{\ln(d)})+O(x^{1+2\ln(d)})) +a(cx^{\ln(d)}d^{\gamma}+O(x^{\ln(d)-1})+O(x^{2\ln(d)}))\\ &=a\ln(x) -cx^{1+\ln(d)}d^{\gamma}+o(1)\\ &\to -\infty\\ \end{array}$

since $1+\ln(d) \approx 0.3 \gt 0$ and $\dfrac{\ln(x)}{x^{1+\ln(d)}} \to 0$.

More precisely, $f(x) \approx \dfrac{x^a}{\exp\left(cx^{1+\ln(d)}d^{\gamma}\right)}(1+o(1))$.

By keeping more terms in the various expansions, we could get a more accurate result.