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Forum: Calculus Topic: Series & Integral 
admin 
Posted: Sat Jun 03, 2017 8:18 pm


Replies: 3 Views: 716

Two more solutions for the first series can be found here. 


Forum: Calculus Topic: $\sum_{n=2}^{\infty}(1)^n \frac{\ln n}{n} =?$ 
admin 
Posted: Sat Jun 03, 2017 7:48 pm


Replies: 2 Views: 643

A solution given by akotronis : Let $\displaystyle{S_{n}:=\sum_{k=1}^{n}(1)^k\frac{\ln k}{k}}$. First observe that, by Dirichlet's criterion, the series converges, because $(1)^k$ has bounded partial sums and $\displaystyle{\frac{\ln k}{k}}$ is eventually decreasing to $0$. Therefore $$\displaysty... 


Forum: Meta Topic: New subforum for Measure and Integration theory 
admin 
Posted: Sat May 20, 2017 1:16 pm


Replies: 1 Views: 388

r9m wrote: Would it be a possible to have a subforum for 'Measure and Integration Theory' that is covered in most graduate classes? Bien sûr! Voila! 


Forum: Calculus Topic: Integral with polylogarithm 
admin 
Posted: Sun Aug 21, 2016 8:28 pm


Replies: 1 Views: 403

A solution by Seraphim We note that $$\zeta(m)  {\rm Li}_m(x)= \sum_{n=1}^{\infty} \frac{1x^n}{n^m}$$ Therefore \begin{align*} S &= \int_{0}^{1}\frac{\zeta(m){\rm Li}_m (x)}{1x} \log^{m1} x \, {\rm d}x \\ &= \sum_{n=1}^{\infty} \frac{1}{n^m} \int_{0}^{1}\frac{1x^n}{1x} \log^{m1} x \... 


Forum: Real Analysis Topic: Absolute convergence criterion 
admin 
Posted: Thu Jul 14, 2016 10:44 am


Replies: 1 Views: 338

A solution may be found here. 


Forum: Real Analysis Topic: \( \int_{0}^{\infty}\frac{x}{\sqrt{e^x1}}\,dx \) 
admin 
Posted: Mon Jul 11, 2016 4:02 pm


Replies: 1 Views: 326

Replied exmember by aziiri : Set \(e^x1=t^2\) then : \[I=\int_0^{\infty} \frac{x}{\sqrt{e^x1}} \ \mathrm{d}x= \int_0^{\infty} \frac{2\ln(t^2+1)}{t^2+1} \ \mathrm{d}t\] Now set \(t=\tan y\) to get : \[I= 4\int_0^{\frac{\pi}{2}} \ln \cos y\ \mathrm{d}y\] The latter integral is widely known to be ... 


Forum: Real Analysis Topic: A nice integral involving sum 
admin 
Posted: Mon Jul 11, 2016 10:03 am


Replies: 6 Views: 894

Replied by exmember aziiri:
Tolaso J Kos wrote: I don't know if induction works, but I believe that your solution can easily be modified for that case also. you just need to substitute \(2014\) with \(n\) in my solution. 


Forum: Real Analysis Topic: A nice integral involving sum 
admin 
Posted: Mon Jul 11, 2016 9:59 am


Replies: 6 Views: 894

Replied by exmember aziiri:
Tolaso J Kos wrote: Azirii, can we generalize that? That is, does the identity $$\int_{\pi}^{\pi}\left ( \sum_{k=1}^{n}\sin \left ( kx \right ) \right )^2\,dx=n\pi$$ hold? Yes, of course. 


Forum: Real Analysis Topic: A nice integral involving sum 
admin 
Posted: Mon Jul 11, 2016 9:57 am


Replies: 6 Views: 894

Replied by exmember aziiri : A possible first step is to use Lagrange's trigonometric identity: \begin{align*} \mathop{\sum}\limits_{k=1}^n {\sin(kx)}&=\frac{1}{2}\frac{\cos\frac{x}{2}\cos\bigl({\bigl({n +\frac{1}{2}}\bigr) x}\bigr)}{\sin\frac{x}{2}}\\ &=........................\\ &=\... 


Forum: Real Analysis Topic: \(\alpha_{n+1}={\sqrt{2}\,}^{\alpha_{n}}\) 
admin 
Posted: Sun Jul 10, 2016 4:09 pm


Replies: 1 Views: 290

Replied by exmember aziiri : \(1<a_n<2\) : we have \(1<a_1<2\). Suppose \(1<a_n<2\) then \(1<{\sqrt{2}\,}^{1}<a_{n+1}={\sqrt{2}\,}^{a_n}<{\sqrt{2}\,}^2=2\). \((a_n)_{n\geq 0} \) is increases. On \([1,2)\) define \(f(x)={\sqrt{2}\,}^xx\) then \(f'(x)= 2^{x/21} \ln 2 1 \leq f'(2)=\ln 2 1 <0\) th... 


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