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Series Contest (SC4)

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Post Sun Aug 03, 2014 3:01 am
Random Variable Integration Guru
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Posts: 381
Integrating by parts I get $$ \begin{align} \int_{0}^{1} \frac{\ln y}{(1-y)^2} \Big(\text{Li}_{2}(y) - \zeta(2) \Big) \ dy &= \int_{0}^{1} \frac{\ln^{2}(1-t)}{t} + \int_{0}^{1} \frac{\ln (t) \ln(1-t)}{1-t} \ dt \\ &= 2 \zeta(3) + \sum_{n=1}^{\infty} \frac{H_{n}}{(n+1)^{2}} \\ &= 2 \zeta(3) + \sum_{n=1}^{\infty} \frac{H_{n+1}}{(n+1)^{2}} - \sum_{n=1}^{\infty} \frac{1}{(n+1)^{3}} \\ &= 2 \zeta(3) + 2 \zeta(3) -1 - \zeta(3) +1 \\ &= 3 \zeta(3) \end{align}$$

You must have done the integration by parts differently.

Post Sun Aug 03, 2014 8:37 am
galactus User avatar
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No, G, that's exactly what I have. It looks like I done the same thing.

disregard that last post with the $Li_{3}$.

I think I messed up a negative sign.

$$\int_{0}^{1}\frac{\log^{2}(1-x)}{x}dx=2\zeta(3)$$

$$\int_{0}^{1}\frac{\log(x)\log(1-x)}{1-x}dx=\zeta(3)$$

$$2\zeta(3)-\zeta(3) = \zeta(3)$$

I had a + between them:

$$\int_{0}^{1}\frac{\log^{2}(1-x)}{x}dx+\int_{0}^{1}\frac{\log(x)\log(1-x)}{1-x}dx=2\zeta(3)+\zeta(3)=3\zeta(3)$$

Heck, I don't know. I know its in there. I have a bigger problem keeping track of negative signs than anything. :roll: :oops:

Post Sun Aug 03, 2014 5:19 pm
Random Variable Integration Guru
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Posts: 381
It should be a plus sign You didn't mess up anything.

Post Sat Sep 06, 2014 4:09 pm
Random Variable Integration Guru
Integration Guru

Posts: 381
Problem 23

$$ \sum_{n=1}^{\infty} \frac{1}{n(16n^2-1)^{2}} = 3 - 3 \log 2 - G$$

Post Sun Sep 07, 2014 1:04 pm
galactus User avatar
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Hey G.

I am sure your method is more efficient and clever, but what I done was to use trigamma and integration to derive the solution.


Break up the summand as $$\sum_{n=1}^{\infty}\frac{1}{(4n-1)^{2}}-\sum_{n=1}^{\infty}\frac{1}{(4n+1)^{2}}-\sum_{n=1}^{\infty}\frac{1}{n(4n-1)(4n+1)}$$

The two let sums are, by definition, the series for trigamma.

That is, $$\sum_{n=1}^{\infty}\frac{1}{(4n-1)^{2}}=1/16\psi_{1}(-1/4)=-G/2+\frac{\pi^{2}}{16}$$

and $$\sum_{n=1}^{\infty}\frac{1}{(4n+1)^{2}}=1/16\psi_{1}(1/4)-1=G/2+\frac{\pi^{2}}{16}-1$$

$$-G/2+\frac{\pi^{2}}{16}-\left(G/2+\frac{\pi^{2}}{16}-1\right)=1-G\tag{1}$$

For the rightmost sum, I just used the geometric series $$\sum_{n=1}^{\infty}x^{4n}=\frac{x^{4}}{1-x^{4}}$$

then integrated and divided by the appropriate x until I hammered it into the required form. You all know what I mean, so I will not write out all the minutae.

Doing the last integration finally gave $$\sum_{n=1}^{\infty}\frac{x^{4n}}{n(4n+1)(4n-1)}=\log(1-x^{2})+\log(1+x^{2})-x\tan^{-1}(x)+z\tanh^{-1}(x)+\frac{\tan^{-1}(x)}{x}+\frac{\tanh^{-1}(x)}{x}$$

$$\lim_{x\to 1}\left[\log(1-x^{2})+\log(1+x^{2})-x\tan^{-1}(x)+z\tanh^{-1}(x)+\frac{\tan^{-1}(x)}{x}+\frac{\tanh^{-1}(x)}{x}\right]=3\log(2)$$

$$\lim_{n\to 0}\left[\log(1-x^{2})+\log(1+x^{2})-x\tan^{-1}(x)+z\tanh^{-1}(x)+\frac{\tan^{-1}(x)}{x}+\frac{\tanh^{-1}(x)}{x}\right]=2$$

Thus, $$3\log(2)-2$$

combine with the other result in (1):

$$1-G-(3\log(2)-2)=3-G-3\log(2)$$

Post Sun Sep 07, 2014 1:57 pm
Random Variable Integration Guru
Integration Guru

Posts: 381
Well, it's not really my approach. It was something that was mentioned in a paper as an alternative way to evaluate some sums that were in one of Ramanujan's first papers.

Simply notice that

$$\psi(1+x) + \psi(1-x) = -2 \gamma - 2x^{2} \sum_{n=1}^{\infty} \frac{1}{n(n^{2}-x^{2})} $$

and

$$ \psi_{1}(1+x) - \psi_{1}(1-x) = - 4x \sum_{n=1}^{\infty} \frac{1}{n(n^{2}-x^{2})} - 4x^{3} \sum_{n=1}^{\infty} \frac{1}{n(n^{2}-x^{2})^{2}} .$$

Then just let $ \displaystyle x= \frac{1}{4}$.

Post Sun Sep 07, 2014 2:20 pm
galactus User avatar
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Yep G, I knew something like that was there but I was too obtuse to see it right off. DUH!!.

Post Mon Aug 24, 2015 8:45 pm
Random Variable Integration Guru
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Posts: 381
While attempting to evaluate something else, I stumbled upon this one:

$$\sum_{k=1}^{\infty} \left(H_{k} - H_{2k} + \log 2 - \frac{1}{4k} \right) = \frac{1}{4} \left(1 - 2 \log 2 \right)$$

I first used summation by parts to determine $$\sum_{k=1}^{n-1} H_{2k} = \left(n- \frac{1}{2} \right)H_{2n} + \frac{H_{n}}{4} - n. $$

EDIT:

I guess you could also evaluate the limit

$$\lim_{x \to 1^{-}} \left[- \frac{\log(1-x)}{1-x} + \frac{1}{2} \left(\frac{\log(1-\sqrt{x})}{1-\sqrt{x}} + \frac{\log(1+\sqrt{x})}{1+\sqrt{x}} \right)+ \log 2 \, \frac{x}{1-x} + \frac{1}{4} \log(1-x)\right] $$

Post Sat Aug 29, 2015 7:46 am

Posts: 50
A similar sum $$S=\sum^\infty_{n=1}\left(H_n-2H_{2n}+H_{4n}-\frac{1}{8n}\right)(1)^{4n}$$
can also be evaluated via a tedious computation of the limit (http://www.wolframalpha.com/input/?i=%2 ... z%20to%201)
\begin{align}
S
&=\lim_{z\to 1}\left[\frac{z^2}{1-z^4}\ln\left(\frac{1-z^2}{1+z^2}\right)-\frac{1}{4}\left(\frac{\ln(1-z)}{1-z}+\frac{\ln(1+z)}{1+z}+\frac{\ln(1+z^2)}{1+z^2}+\frac{2z\arctan(z)}{1+z^2}\right)+\frac{\ln(1-z^4)}{8}\right]\\
&=\frac{1}{8}-\frac{\pi}{16}
\end{align}
or by using summation by parts as Random Variable suggested. As of now, I am not sure how to handle the squared version of the sum though.

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