Search found 36 matches
- Sun Dec 06, 2015 10:00 am
- Forum: Real Analysis
- Topic: Estimation of a sum with cosecants
- Replies: 2
- Views: 2602
Re: Estimation of a sum with cosecants
Thank you for the answer Demetres. The problem is from here.
- Sun Dec 06, 2015 9:55 am
- Forum: Real Analysis
- Topic: Estimation of a sum with cosecants
- Replies: 2
- Views: 2602
Estimation of a sum with cosecants
Show that $$\sum_{k=1}^{n-1}\csc\left(\frac{k\pi}{n}\right)=\frac{2}{\pi}n\ln n+\left(\frac{2\gamma}{\pi}-\frac{2\ln(\pi/2)}{\pi}\right)n+\mathcal O(1)\,.$$
- Sun Nov 15, 2015 5:30 pm
- Forum: Real Analysis
- Topic: Series involving Riemann zeta function
- Replies: 2
- Views: 2636
Re: Series involving Riemann zeta function
We have \(\begin{align*}\displaystyle\sum_{n\geq2}\frac{(-1)^n\zeta(n)}{n(n+1)}=\sum_{n\geq2}\frac{(-1)^n\zeta(n)}{n}-\sum_{n\geq2}\frac{(-1)^n\zeta(n)}{n+1}:=A-B\end{align*}\). Now from the Taylor series of \(\ln(1-x)\) it is \(\begin{align}\label{1}\displaystyle\sum_{n\geq2}\frac{x^n}{n}&=-\ln...
- Sun Nov 15, 2015 4:38 pm
- Forum: Real Analysis
- Topic: Integral involving \(\Gamma\) function
- Replies: 4
- Views: 4125
Re: Integral involving \(\Gamma\) function
Is there a "direct" way computing of $\displaystyle\sum_{n=2}^\infty \frac{(-1)^n\,\zeta(n)}{n\,(n+1)}$ ?
- Sun Nov 15, 2015 11:32 am
- Forum: Real Analysis
- Topic: \(\sum_{n=1}^{+\infty}({-1})^{n+1}\frac{\sin{n}}{n}\)
- Replies: 4
- Views: 3971
Re: \(\sum_{n=1}^{+\infty}({-1})^{n+1}\frac{\sin{n}}{n}\)
A nice relevant problem, proposed by Ovidiu Furdui, that is still open for submitting solutions is H722 here
http://www.fq.math.ca/Problems/2012AugustAdvanced.pdf
http://www.fq.math.ca/Problems/2012AugustAdvanced.pdf
- Sat Nov 14, 2015 3:10 am
- Forum: Real Analysis
- Topic: \(\sum(-1)^{n}(\tfrac{1}{\sqrt{n}}+\tfrac{(-1)^{n-1}}{n})\)
- Replies: 1
- Views: 1930
Re: \(\sum(-1)^n(\frac{1}{\sqrt{n}}+\frac{(-1)^{n-1}}{n})\)
Set \(S_N:=\sum_{n=1}^{N}(-1)^n\left(\frac{1}{\sqrt{n}}+\frac{(-1)^{n-1}}{n}\right)\), so that \(S_N=-H_N+\sum_{n=1}^{N}\frac{(-1)^n}{\sqrt{n}}\). By Dirichlet's criterion \(\sum_{n\geq1}\frac{(-1)^n}{\sqrt{n}}\) converges, so, since \(H_N=\ln N+\mathcal O(1)\) we have \(S_N=-\ln N+\mathcal O(1)\) t...
- Wed Nov 11, 2015 1:46 pm
- Forum: Real Analysis
- Topic: $\int_{0}^{+\infty}{\frac{\log({\frac{1}{x}})}{(1+x)^n}dx}$
- Replies: 3
- Views: 3094
Re: $\int_0^{\infty}\frac{\log({\frac{1}{x}})}{(1+x)^n}dx$
Another solution: We set \(\displaystyle{a_n:=-\int_{0}^{+\infty}\frac{\ln x}{(1+x)^n}\,dx}\), multiply with \(y^n\) and sum for \(n\geq2\). For \(y\in(-1,0)\cup(0,1)\) we will have: \[\begin{aligned}\sum_{n\geq2}a_ny^n&=-\sum_{n\geq2}\left(\int_{0}^{+\infty}\frac{\ln x}{(1+x)^n}\,dx\right)y^n \...
- Tue Nov 10, 2015 10:47 pm
- Forum: Real Analysis
- Topic: A (combinatorial..?) identity
- Replies: 2
- Views: 2501
Re: A (combinatorial..?) identity
Thank you for your solution Demetres. I don't have a combinatorial proof either. Here is the solution I had given for this one on http://www.mathematica.gr.: At first we have \(\begin{aligned}\displaystyle\sum_{k=0}^{n-1}(-1)^{n-k-1}\dfrac{(n+k)!}{(k!)^2(n-k-1)!}&=\sum_{k=0}^{n-1}(-1)^{n-k-1}(n-...
- Tue Nov 10, 2015 10:13 pm
- Forum: Real Analysis
- Topic: Recurrent sequence
- Replies: 0
- Views: 1831
Recurrent sequence
Let \(a_n\) be defined by \(\displaystyle{a_n=n(n-1)a_{n-1}+\frac{n(n-1)^2}{2}a_{n-2}}\) for \(n\geq3\) and \(a_{1}=0,\;a_2=1\). \(\displaystyle{1)}\) Show that \(\displaystyle{\lim_{n\to+\infty}\frac{e^{2n}a_n}{n^{2n+1/2}}=2\sqrt{\frac{\pi}{e}}}\) and \(\displaystyle{2)}\) compute \(\displaystyle{\...
- Tue Nov 10, 2015 10:12 pm
- Forum: Real Analysis
- Topic: Estimation of a parametric integral
- Replies: 0
- Views: 1768
Estimation of a parametric integral
Evaluate the following limits if they exist
1) \(\displaystyle \lim_{s\to+\infty}\frac{1}{\ln s}\int_{0}^{+\infty}\frac{e^{-\frac{x}{s}-\frac{1}{x}}}{x}\,dx\)
2) \(\displaystyle \lim_{s\to+\infty}\left(\int_{0}^{+\infty}\frac{e^{-\frac{x}{s}-\frac{1}{x}}}{x}\,dx-\ln s\right)\).
1) \(\displaystyle \lim_{s\to+\infty}\frac{1}{\ln s}\int_{0}^{+\infty}\frac{e^{-\frac{x}{s}-\frac{1}{x}}}{x}\,dx\)
2) \(\displaystyle \lim_{s\to+\infty}\left(\int_{0}^{+\infty}\frac{e^{-\frac{x}{s}-\frac{1}{x}}}{x}\,dx-\ln s\right)\).