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2 votes
4 answers
270 views

Show that $\lim_{n\to\infty}n\left(n\ln{n}+\ln{\sqrt{2}}-n-\sum_{k=1}^n \ln{\left(k-\frac{1}{2}\right)}\right)=\frac{1}{24}$.

I am trying to show that $$L=\lim\limits_{n\to\infty}n\left(n\ln{n}+\ln{\sqrt{2}}-n-\sum\limits_{k=1}^n \ln{\left(k-\frac{1}{2}\right)}\right)=\frac{1}{24}$$ Desmos strongly suggests that this is true,...
Dan's user avatar
  • 25.7k
5 votes
2 answers
134 views

Find $C$ such that $\frac{1}{n}\prod_{k=1}^{n}C\left(\cos{\frac{k\pi}{2(n+1)}}+\sin{\frac{k\pi}{2(n+1)}}-1\right)$ converges to a positive number.

I'm looking for the value of $C$ such that $L=\lim\limits_{n\to\infty}\frac{1}{n}\prod\limits_{k=1}^{n}C\left(\cos{\frac{k\pi}{2(n+1)}}+\sin{\frac{k\pi}{2(n+1)}}-1\right)$ equals a positive real ...
Dan's user avatar
  • 25.7k
5 votes
0 answers
131 views

Infinite product of areas in a square, inscribed quarter-circle and line segments.

The diagram shows a square of area $An$ and an enclosed quarter-circle. Line segments are drawn from the bottom-left vertex to points that are equally spaced along the quarter-circle. The regions ...
Dan's user avatar
  • 25.7k
3 votes
0 answers
121 views

Special property of circle with radius 0.975399...

$4n$ points are uniformly distributed on a circle. Parabolas are drawn in the manner shown below with example $n=4$. The parabolas' vertices are at the center of the circle. The parabolas have a ...
Dan's user avatar
  • 25.7k
6 votes
2 answers
270 views

What is a closed form of this limit? (product of areas in circle with parabolas)

I am looking for a closed form of $L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^n \left(\left(6+\frac{4n}{\pi}\left(\sin{\frac{\pi}{2n}}\right)\left(\cos{\frac{(2k-1)\pi}{2n}}\right)\right)^2-\left(\...
Dan's user avatar
  • 25.7k
0 votes
1 answer
54 views

Find $a$ such that the limit is zero

Problem : Let $x>0$ then define : $$f(x)=\left(\left(\frac{1}{x}\right)!\left(x!\right)\right)^{\frac{1}{x+\frac{1}{x}}}$$ Then find $a$ such that : $$\lim_{x\to\infty}f(x)-\frac{1}{2}\left(\frac{1}...
Ranger-of-trente-deux-glands's user avatar
16 votes
1 answer
453 views

Closed form for $\lim\limits_{n\to\infty}\prod\limits_{k=1}^n{\left(2-\frac{2n^2-\pi^2+8}{n^2}\cos{\frac{(2k-1)\pi}{n}}\right)}$?

I am looking for a closed form for: $$\lim_{n\to\infty}\prod_{k=1}^n{\left(2-\frac{2n^2-\pi^2+8}{n^2}\cos{\frac{(2k-1)\pi}{n}}\right)}$$ (Wolfram suggests that it's approximately 6.17966.) Context: I ...
Dan's user avatar
  • 25.7k
1 vote
1 answer
61 views

A Regularized Beta function limit: $\lim_{a\to0}\frac{1-\text I_\frac zb(a,b)}a$

The goal is to “generalize” the Exponential Integral $\text{Ei}(x)$ using the Regularized Beta function $\text I_z(a,b)$: $$f(b,z)=\lim_{a\to0}\frac{1-\text I_\frac zb(a,b)}a$$ Some clues include: $$\...
Тyma Gaidash's user avatar
34 votes
0 answers
596 views

An iterative logarithmic transformation of a power series

Consider the following iterative process. We start with the function having all $1$'s in its Taylor series expansion: $$f_0(x)=\frac1{1-x}=1+x+x^2+x^3+x^4+O\left(x^5\right).\tag1$$ Then, at each step ...
Vladimir Reshetnikov's user avatar
2 votes
1 answer
91 views

Closed form of the series: $\sum _{n=0}^{\infty }\left(\frac{1}{2^n\left(1+\sqrt[2^n]{x}\right)}\right)$

Came across this in a calc textbook from the 1800s and I can't figure out a way to solve it. Trying to write it in product form by taking the integral didn't work. I also tried adding consecutive ...
isty43's user avatar
  • 113
48 votes
1 answer
1k views

How to evaluate double limit of multifactorial $\lim\limits_{k\to\infty}\lim\limits_{n\to 0} \sqrt[n]{n\underbrace{!!!!\cdots!}_{k\,\text{times}}}$

Define the multifactorial function $$n!^{(k)}=n(n-k)(n-2k)\cdots$$ where the product extends to the least positive integer of $n$ modulo $k$. In this answer, I derived one of several analytic ...
TheSimpliFire's user avatar
  • 27.1k
7 votes
1 answer
342 views

An infinite product for $\frac{\pi}{2}$

Please help prove $$ \begin{align} \frac{\pi}{2}&=\left(\frac{1}{2}\right)^{2/1}\left(\frac{2^{2}}{1^{1}}\right)^{4/(1\cdot 3)}\left(\frac{1}{4}\right)^{2/3}\left(\frac{2^{2}\cdot4^{4}}{1^{1}\...
tyobrien's user avatar
  • 3,557
5 votes
1 answer
254 views

Evaluate $\lim\limits_{n\to\infty} \frac{\sin(1)+\sin^2(\frac{1}{2})+\ldots+\sin^n(\frac{1}{n})}{\frac{1}{1!}+\frac{1}{2!}+\ldots+\frac{1}{n!}}$

This was a recent problem on the Awesome Math Problem Column. The solution is given as follows: We shall use Stolz-Cesaro Lemma. We have: $$\lim_{n\to\infty} \frac{\sin(1)+\sin^2(\frac{1}{2})+\ldots+\...
drfrankie's user avatar
  • 519
8 votes
1 answer
326 views

Challenging limit: $\lim_{\alpha\to0^{+}}\left(\frac{1}{2\alpha}-\int_1^\infty\frac{dx}{\sinh(\pi\alpha x)\sqrt{x^2-1}}\right)$

Here is a challenging limit proposed by a friend: $$\lim_{\alpha\to0^{+}}\left(\frac{1}{2\alpha}-\int_1^\infty\frac{dx}{\sinh(\pi\alpha x)\sqrt{x^2-1}}\right)$$ and he claims that the closed form ...
Ali Shadhar's user avatar
  • 25.8k
1 vote
2 answers
158 views

Finding a closed form to a minimum of a function

It's a try to find a closed form to the minimum of the function : Let $0<x<1$ then define : $$g(x)=x^{2(1-x)}+(1-x)^{2x}$$ Denotes $x_0$ the abscissa of the minimum . Miraculously using Slater's ...
Ranger-of-trente-deux-glands's user avatar

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