Hey Z man:). Here is two other interesting Euler-like sums that do converge. What are your thoughts on finding a closed form?.
$$\sum_{n=1}^{\infty}\left[-(H_{n}^{(2)})^{3}+\frac{\pi^{2}}{2}(H_{n}^{(2)})^{2}-\frac{\pi^{4}}{12}H_{n}^{(2)}+\frac{\pi^{6}}{216}\right]...[1]$$
and/or...
Here is another to consider, Z. This converges to $2\zeta(3)-\zeta(2)$
$$\sum_{n=1}^{\infty}H_{n}\left(\zeta(3)-H_{n}^{(3)}\right)=-1/2\sum_{n=1}^{\infty}H_{n}\psi^{(2)}(n+1)$$
Hey Z:
Here is another perhaps we can find a gen. func for. I feel I should already have this one.
$$\sum_{n=1}^{\infty}(H_{n})^{3}y^{n}$$
I derived several fun sums by successive diffing of the digamma.
If we note: $$G(x)=\frac{\Gamma(x)n!}{\Gamma(n+x+1)}$$
and...
Cool thread, Z.
Here are a few thoughts. I am going to use $$\psi_{1}(n+1)$$ simply for ease of computations. We can adjust it for $\psi_{1}(n)$.
Attempt to find a gen. func. for $$\sum_{n=1}^{\infty}\left(\psi_{1}(n+1)\right)^{2}y^{n}$$
Note that:
$$\psi_{1}(n+1)=\zeta(2)-H_{n}^{(2)}$$...
Hello All.
It has been while since I posted here. I hope no one minds I give my two cents. Since I noticed Pranav frequents this site, I thought I would say howdy.
Anyway, something I often use when confronted with a sum like this is to use the digamma function.
Note the classic...
I made a sub $t=\frac{\pi x}{2}$ and arrived at
$$\frac{1}{2\pi}\int_{0}^{\infty}\frac{1}{(t^{2}+1)cosh(\frac{\pi t}{2})}$$
There is an identity that says $$\frac{1}{2\pi}\int_{0}^{\infty}\frac{1}{(t^{2}+1)cosh(at)}dt=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2a+(2n-1)\pi}$$.
It is derived from...
Hi All:
I am new to the site, so I thought this would be a good time to post an interesting integral I ran across that I am having a time with. It is a miscellaneous problem in Schaum's Outline of Complex Variables, #86 in ch. 7. I have been self-teaching a little CA when I get time and this...
Hi RV. Cool problem.$$\frac{\pi}{4}-\tan^{-1}(1/3)+tan^{-1}(1/5)-\tan^{-1}(1/7)+\cdot\cdot\cdot$$
$$\left(\frac{\pi}{4}+\tan^{-1}(1/5)+\tan^{-1}(1/9)+\cdot\cdot\cdot\right) -\left(\tan^{-1}(1/3)+tan^{-1}(1/7)+\tan^{-1}(1/11)+\cdot\cdot\cdot \right)$$
This can now be written as...
It can be shown with some clever maneuvering and the use of the Gamma function that:
\int_{0}^{\infty}\frac{sin(x)}{x^{a}}dx=\frac{\sqrt{\pi}{\Gamma}(1-\frac{a}{2})}{{\Gamma}(\frac{a}{2}+\frac{1}{2})}
Gamma is undefined at 0, so one can see that a=2 leads to Gamma(0) and a=0 gives 1.
Of...
I was pondering this integral a little today. I see from the MathWorld site that the solution is
\frac{\text{Catalan}}{2}+\frac{\pi}{8}ln(2)
I think if we are clever enough, we may be able to transform the integral into some other integrals that can be done. Though not by elementary means...