@TedShifrin yes... when I was walking my dog this morning, there were two bunnies in the park. My son and I caught them (or coyotes would by tonight) and we spent the day getting a cage and food and stuff for them
@Chris'ssis I read on the internet that it's the real life thing closest to the NZT48 (as in the limitless movie :P) .. but it doesn't feel like that .. I just don't feel like sleeping and eating while the effect lasts (that's it)
@Chris'ssis modafinil has unknown long term effect (being a relatively recent drug in the market .. so I can't take the risk of keeping on using them ! .. )
One of the possible ways of computing the series is to obtain the generating function, but
this might be a tedious, hard work, pretty hard to achieve. What would you propose then?
$$\sum_{n=1}^{\infty} \frac{H_{
n}}{2^nn^4}$$
@Chris'ssis omg ! with generating functions I can't even dream of trying it .. $\sum_{n=1}^{\infty} \frac{H_{n}}{2^nn^3}$ almost killed me (while I was reading the answers in main) ..
In the same spirit as Robert Israel's answer and continuing Raymond Manzoni's answer (both of them deserve the credit because of inspiring my answer) we have
$$
\sum_{n=1}^\infty \frac{H_nx^n}{n^2}=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operat...
@r9m That one is wrong. Yesterday I computed the correct variant.
@Chris'ssis there are some stuff that I got stuck on once I took a look at his answer and tried to see if I could do it on my own =_= some non trivial stuff must have been used without a proof/mentioning
If it is not impolite to ask, could you have a look at this question? [Two ways to evaluate ∫(Δu)vdΩ, two different results](http://math.stackexchange.com/questions/933579/two-ways-to-evaluate-int-delta-u-v-d-omega-two-different-results)
Start with integration by parts (IBP) by setting $u=\ln^3(1+x)$ and $dv=\dfrac{\ln x}{x}\ dx$ yields
\begin{align}
\int_0^1\frac{\ln^3(1+x)\ln x}x\ dx&=-\frac32\int_0^1\frac{\ln^2(1+x)\ln^2 x}{1+x}\ dx\\
&=-\frac32\int_1^2\frac{\ln^2x\ln^2 (x-1)}{x}\ dx\quad\Rightarrow\quad\color{red}{x\mapsto1+x...
But I have nothing against, some say it's an epic answer. :-)
@BalarkaSen $x=1/2$ produces some magic cancellations. :-) When you use $x=1/3$ you realize immediately the generating function is not correct, you get a different numerical result.
@BalarkaSen No need for worshiping here. One of the worst things that can happen to someone is to be surrounded by people that apparently admire you, but they don't give a s**t on you. Maybe they are only interested in taking some advantage on your back.:-)
@Chris'ssis I used to like two girls who were good at running and were very beautiful. Since you like running, I think you must be very beautiful too? =)
@Chris'ssis You're correct. He's made a fatal mistake. I've edited his answer but only in the parts that I can locate his mistake. I hope my edited version is correct
In the same spirit as Robert Israel's answer and continuing Raymond Manzoni's answer (both of them deserve the credit because of inspiring my answer) we have
$$
\sum_{n=1}^\infty \frac{H_nx^n}{n^2}=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operat...
@Anastasiya-Romanova look at the variable change in $(2)$
Once you do that, how can you recover what is added or lost by that operation? That constant he tried to get later isn't stable, it changes while you plug different value in the generating function.
@Chris'ssis Yeah, you're correct. He is lucky get a correct answer since it is evaluated at $\frac{1}{2}$, so changing variable by $x\mapsto1-x$ doesn't effect at all
@Anastasiya-Romanova Actually, I'd go further and say it works because he fits the proper constant there by I don't know what means. :-)
@Anastasiya-Romanova His initial answer was $$\sum_{n=1}^\infty \frac{H_n}{2^nn^3}=\color{purple}{\frac{7\pi^4}{720}+\frac{\ln^42}{24}-\frac{\ln2}8\zeta(3)+\operatorname{Li}_4\left(\frac12\right)}$$
and then he magically gets rid of that $7$ in front of the first fraction.
I love Random Variable's comment there
"Even though you seem to have made a minor error somewhere (which I'm unable to locate), your answer is impressive and deserves an upvote. +1 – Random Variable"
@Anastasiya-Romanova I really appreciate Tunk Fey, really! Unfortunately he did a ugly mistake there that continued further.
@robjohn I think I'm going to offer a 500 points bounty for this question. I'm curious to see if someone can get the precise value of the limit, but, of course, I'm very very curious to see what that limit is.
In the chatroom we discussed about the asymptotic of $\displaystyle \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}$, and if we think of the inverse tangent integral, it's easy to see that $\displaystyle \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}\approx \operatorname{Ti}_2(n)\approx \frac{\pi}{2...
@WillHunting: I didn't say YOU were naive. I personally find it naive, if someone idolises something as much as you do whilst never having experienced it before. Not the person - the action.
@Huy No problem. It's alright to say I am naive. What I mean is this: if option A is so bad, option B is very inviting even if you never had B before...
@WillHunting: Personally, I'd rather experience both and then make a choice instead of fully committing to something I've never done before - the latter being what seemed to me you were doing.
@Huy Well, another reason is that there are some things I am certain option B has but A does not, and those things are very important to me, though they may not be to others.
@Chris'ssis Good catch. I didn't notice that. I assumed that the error was indeed just the integration constant and I didn't bother to check a second time.