I was just looking at an answer and it seems that no one noticed something important there.
$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\left(\frac{\pi^2}{6} \quad -\sum_{k=1}^{n}\frac{1}{k^2}\,\right)=\frac{\pi^2}{4}\ln 2 -\zeta(3)$$ This is not explained at all.
@robjohn @r9m I wonder if this one can be awesomely done by series manipulations only :-) $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n }\left(1+\frac{1}{2^2}+\cdots +\frac{1}{n^2}\right)$$
I feel like I am going to leave this degree with nice knolwedge in probability, combinatorics, graphs, discrete mathematics, logic and set theory and all that. But my analysis and calculus are just high-school level with a bit more experience. I really some good book for these two.
sometimes I hate texmaker. I was crying like crazy earlier because I pressed save so many times and the file rolled back to only the first chapter being typed T___T!
We will make frequent use of
$$
\binom{n+1}{k+1}=\binom{n}{k}\frac{n+1}{k+1}\tag{1}
$$
The Generalized Harmonic Numbers of the second order are defined as
$$
H_n^{(2)}=\sum_{k=1}^n\frac1{k^2}\tag{2}
$$
The factor of $2^{-n}$ in each term reminded me of the Euler Series Transformation. Reversing t...
Let $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$
where $H^{(p)}_n = \sum_{i=1}^n i^{-p}$, the $n$th $p$-harmonic number. The $A(p,q)$'s are known as alternating Euler sums.
Can someone provide a nice proof that
$$A(1,1) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} =...
@usukidoll: I'm in XII grade, i do fun stuff like the above in my free time. I should actually be doing inverse trigonometry, but I can't remember silly conditions :(
@Alizter: Some people like to argue that the asymptotes are tangents which touch the function at $\infty$. Now, these are the same people that say that two parallel lines intersect each other at infinite distance. :|
@Chris'ssis Incredible !!! :D I am still struggling with it :( .. oh $H_n^{(2)-}$ stands for the alternating sum $\sum\limits_{k=1}^n \frac{(-1)^k}{k^2}$
@Alizter: I'm making pseudo statements in my attempt to find the next "I am a liar" phrase. It would be wise to ignore the non-mathematical things I say.
Let $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$
where $H^{(p)}_n = \sum_{i=1}^n i^{-p}$, the $n$th $p$-harmonic number. The $A(p,q)$'s are known as alternating Euler sums.
Can someone provide a nice proof that
$$A(1,1) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} =...
