18
votes
Do roots of these polynomials approach the negative of the Euler-Mascheroni constant?
The below answer consists of two parts. At first, we prove that each $p_n(z)-z$ has the unique negative root. Next, we describe the limit as the negative root of a certain function.
Denote $x_k=k(k-1)/...
- 109k
16
votes
Accepted
Another limit involving the fractional part
The conjecture is true. For $t\in[0,1]$, let $N(t)$ be the number of
$k\in\{1,2,\dotsc,n\}$ satisfying $\{n/k\}<t$. On the one hand,
$$\sum_{k=1}^n f\left(\frac{n}{k}\right)=\int_0^{1/2}\bigl(N(t+1/...
- 105k
12
votes
Do roots of these polynomials approach the negative of the Euler-Mascheroni constant?
In this answer, I'll give an explicit formula for the interpolating polynomial and their limit. As a result, I'll conclude that
(1) $p_k(x)-x$ is increasing as a function of $x$ on $(-\infty, 0]$, ...
- 156k
8
votes
Do roots of these polynomials approach the negative of the Euler-Mascheroni constant?
According to Mathematica your polynomials satisfy the recurrence relation
$$
(2 n+1) p(n) \left(n^2+3 n-2 x+2\right)+p(n+1) \left(-4 n^3-18 n^2+4 n x-27 n+2 x-14\right)+(2 n+3) (n+2)^2 p(n+2)=0
$$
...
4
votes
Accepted
Beta function, harmonic numbers, and integral values
TL;DR
Peter Taylor pointed out that the expressions below reduce to
$$\lim_{n\rightarrow k}x^{-n}I_n(x)= (x-1) \sum_{i=0}^k \frac{H_{k+1} - H_i}{x^{i+1}} -2 \,\text{arctanh}\,(1-2 x)$$
$\newcommand\...
- 188k
4
votes
Accepted
Proof of a zeta function limit
Use the expansion
$$\zeta(s)=\frac{1}{s-1}+\gamma+{\cal O}(s-1),$$
hence, since $\zeta(x)=1+2^{-x}+3^{-x}+4^{-x}+5^{-x}+\cdots$, you have
$$\zeta(\zeta(x))=\frac{2^x}{1+(2/3)^x+(2/4)^x+(2/5)^x+\cdots}+...
- 188k
2
votes
Another limit involving the fractional part
Here are more details on my comment. Let's admit that we have
$$(1)\,\,\,\, \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{n}{k}\right)=\int_{0}^ {1}f\left(\frac{1}{t}\right)dt$$
Since ...
- 611
1
vote
A set of divergent integrals that I think, equal to $-\gamma$
There might be a mistake here, but formally I'm getting
\begin{align}
\int_{x=0}^{x=\infty} \frac{\mathrm{sgn}(x-1)}{x} \mathrm{d}x
&= \int_{x=1}^{x=\infty} \frac{1}{x} \mathrm{d}{x} - \int_{x=0}^{...
- 2,203
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