Reputation
34,222
Next tag badge:
267/100 score
19/20 answers
Badges
6 101 189
Newest
 Enlightened
Impact
~1.1m people reached

Jun
9
comment Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$
How do you go from $\int_0^1 \frac{1-t^k}{1-t} \, dt$ to $\int_0^1 \ln(1-t) (-kt^{k-1}) \, dt$ in the second step?
May
29
awarded  Nice Answer
May
18
awarded  Nice Question
Apr
17
awarded  Announcer
Mar
24
awarded  Announcer
Mar
20
awarded  Enlightened
Mar
20
awarded  Nice Answer
Mar
5
awarded  Announcer
Feb
24
awarded  Notable Question
Feb
22
awarded  Announcer
Feb
20
awarded  Nice Answer
Feb
16
awarded  Enlightened
Feb
16
awarded  Nice Answer
Feb
8
awarded  Nice Answer
Jan
19
awarded  Necromancer
Jan
13
comment Does exceptionalism persist as sample size gets large?
I had to do the transformation by hand to verify it, but you are correct. Nice observation! For the record, the Mudholkar, Chaubey, and Tian paper does not mention that $\log Z - \log Y$ has that simpler form.
Jan
2
revised Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$
improved formatting
Jan
2
comment Double Euler sum $ \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} $
The Borwein and Girgensohn paper proves that the answer to your question is "Yes" for certain values of $a$, $b$, and $c$. That paper is nearly 20 years old, though, and the state of the art may have improved since then.
Jan
2
comment Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$
I just spent two hours typesetting this, and I'm tired of looking at it. If you spot any typos, let me know, and I'll fix them when I get the chance.
Jan
2
answered Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$