Linked Questions

0
votes
0answers
80 views

Limit Ratio of Gamma Functions [duplicate]

How can one prove that for any $x>1$: $$\lim\limits_{n\to \infty }\left(\frac{n^x \Gamma \left(n+1\right)}{\Gamma \left(x+n+1\right)}\right)=1$$ It is easy to show this for $x$ a natural number ...
19
votes
5answers
2k views

Evaluate $\int_0^\infty\frac{1-e^{-x}(1+x )}{x(e^{x}-1)(e^{x}+e^{-x})}dx$

\begin{equation} \int_0^\infty\frac{1-e^{-x}(1+x )}{x(e^{x}-1)(e^{x}+e^{-x})}dx \end{equation} My colleague got this problem from his friend but he didn't know the answer so he asked my help. ...
33
votes
3answers
1k views

Prove $\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$

I am trying to prove that $$\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$$ I know how to deal with integrals ...
16
votes
6answers
696 views

A closed form for the infinite series $\sum_{n=1}^\infty (-1)^{n+1}\arctan \left( \frac 1 n \right)$

It is known that $$\sum_{n=1}^{\infty} \arctan \left(\frac{1}{n^{2}} \right) = \frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})}\right). $$ Can we also find ...
17
votes
7answers
772 views

Prove the existence of limit of $x_{n+1}=x_n+\dfrac{x_n^2}{n^2}$

The problem is: Let $\{x_n\}$ be a sequence such that $0<x_1<1$ and $x_{n+1}=x_n+\dfrac{x_n^2}{n^2}$. Prove that there exists the limit of $\{x_n\}$. It is easy to show that $x_n$ is increasing,...
14
votes
2answers
1k views

How to find this infinite product

How to find this infinite product ? $$\prod_{n=0}^\infty \left(1-\dfrac{2}{4(2n+1)^2+1}\right)$$ I try to use infinite product of $\cos{x}$ but it doesn't work. Thank you.
11
votes
7answers
1k views

A gamma function inequality

I would like to prove $$\frac{\Gamma(n+\frac{1}{2})}{\Gamma(n+1)} \le \frac{1}{\sqrt{n}}$$ for all natural $n \ge 1$. The inequality does seem to be true numerically, but the proof eludes me.
16
votes
2answers
3k views

$\int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2)$ How to show?

$$ \int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2) $$ Anyone an idea on how to prove this?
15
votes
3answers
930 views

challenging alternating infinite series involving $\ln$

I ran across an infinite series that is allegedly from a Chinese math contest. Evaluate: $\displaystyle\sum_{n=2}^{\infty}(-1)^{n}\ln\left(1-\frac{1}{n(n-1)}\right).$ I thought perhaps this ...
13
votes
5answers
692 views

Closed form for an infinite product

The following fascinating formula appears in the paper "On gamma quotients and infinite products" by M.Chamberland and A.Straub (see page 9): $$\prod_{n=2}^\infty\left(1-\frac{1}{n(n-1)}\right)=-\...
6
votes
8answers
296 views

Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$

I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\...
15
votes
2answers
321 views

Need help with $\int_0^1\frac{\log(1+x)-\log(1-x)}{\left(1+\log^2x\right)x}\,dx$

Please help me to evaluate this integral $$\int_0^1\frac{\log(1+x)-\log(1-x)}{\left(1+\log^2x\right)x}\,dx$$ I tried a change of variable $x=\tanh z$, that transforms it into the form $$\int_0^\infty\...
10
votes
4answers
283 views

Evaluating $-\int_0^1\frac{1-x}{(1-x+x^2)\log x}\,dx$

I was trying do variations of an integral representation for $\log\frac{\pi}{2}$ due to Jonathan Sondow, when I am wondering about if it is possible to evaluate $$\int_0^1-\frac{1-x}{(1-x+x^2)\log ...
17
votes
1answer
6k views

Quotient of gamma functions?

I'm sorry if this is a simple question, but this page on Wolfram Research states that it follows from Stirling's formula that: $$ \frac{\Gamma(x+\beta)}{\Gamma(x)} \approx x^\beta $$ for large $x$, ...
6
votes
4answers
591 views

Why do the endpoints of the Maclaurin series for arcsin converge?

The series $$\sum_{n=0}^\infty {{-\frac {1} 2} \choose n} \frac{(-1)^n}{2n+1}$$ is an endpoint for the Maclaurin series for arcsin(x). (The other endpoint is just the negative of this one.) I played ...

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