series sum

[man111]

[tex]\displaystyle \sum^{\infty}_{k=1}\frac{1}{k^22^k}[/tex]

[Sup3rlum]

Така като го гледам това e $Li_{2}(1/2) = \frac{\pi^2}{12}-\frac{ln^2(2)}{2}$

Не съм много сигурен как трябва да се изчисли на ръка, но ето два варианта ако имаш, така да кажем, "Инструменти"

Нека разгледаме функцията редицата:

$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1} =x \sum_{n=0}^{\infty}\frac{x^{n}}{n+1}$

$-\frac{ln(1-x)}{x} =\sum_{n=0}^{\infty}\frac{x^{n}}{n+1}$

Интегрираме:

$Li_{2}(x) = \sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)^2}$
$Li_{2}(x) = \sum_{n=1}^{\infty}\frac{x^n}{n^2}$
$Li_{2}(1/2) = \sum_{n=1}^{\infty}\frac{1}{2^nn^2}$

По принцип знам че $Li_{a}(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^a}$ От където излиза нашата сумация, $a=2, x=\frac{1}{2}$ но си нямам идея как се изчислява без $Li$ , то нали за това е направен, защото няма еквивалентен израз в затворена форма с елементарни функции. Може би някой по знаещ ще каже това как се намира в задача състезание 9-12 клас.


Едит: Набарах едно айдентити, ама не знам кой на тая планета ще се сети по време на състезание:


$\sum_{n=1}^{\infty}\frac{1}{2^nn^2} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{\frac{1}{2}^n}{n^2} + \frac{\frac{1}{2}^n}{n^2} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{\frac{1}{2}^n}{n^2} + \sum_{n=1}^{\infty}\frac{(1-\frac{1}{2})^n}{n^2} = \frac{1}{2}(Li_2(\frac{1}{2})+Li_2(1-\frac{1}{2}))$

Ако знаем това:
$Li_2(x) + Li_2(1-x) = \frac{\pi^2}{6} - ln(x)ln(x-1)$
Заместваме 1/2 и получаваме отговора. Някви по-смислени методи?

[man111]

please explain me how u get last line. Thanks

[Sup3rlum]

please explain me how u get last line. Thanks

Hey, didn't know you weren't Bulgarian. First I'd like you to know that in this forum (and like in most other Bulgarian forums) it is strictly necessary that we write in Bulgarian using Cyrillic, so do not be surprised if you get offered no help

With that said, on to the math question:

What you have posted initially is actually the series expansion of a non-elementary function called $Li_2(x)$ at $x=1/2$ . In order to get out a closed form value out of that sum you need to: either derive some of the properties of $Li_2$ or straight up know them. In my previous response I have described how I have split the sum into two parts which fall nicely into an identity, that I've put at the end of the post. You could also go straight up go and derive that identity using the sums I have posted initially.

By the way, this section is called "Math competitions and/or Olympiads , 9th-12th grade"
I'm not sure I know of any smart trick to compute this under timed conditions.

Have a nice day.