Question

haha for fun :P
find the exact value of
\(S=1+\frac{1}{2^6}+\frac{1}{3^6}+\frac{1}{4^6}+...\)

Answer 1

\(\large\tt \begin{align} \color{black}{
\dfrac{\pi^6}{945}\hspace{.33em}\\~\\

}\end{align}\)

Answer 2

yes , but why :P

Answer 3

lol, wolfram gave it

Answer 4

:P ohk lol show a method

Answer 5

i only know 2 method, so looking for more excited answers

Answer 6

http://www.wolframalpha.com/input/?i=%5Csum_%7Bi%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7Bi%5E6%7D

Answer 7

@AkashdeepDeb do u memorize this ?
what function for Taylor does this represent ! (sorry not convinced enough )

Answer 8

Well since \[\Large \frac{\pi^2}{6} = \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...\] and we can rewrite what you put as \[\Large \frac{1}{1^2} + \frac{1}{8^2}+\frac{1}{27^2}+\frac{1}{64^2}+...\] I can at least put the most amount of reasoning into an upper bound as pi^2/6 which at least has a method to it. =P

Answer 9

i see well , same method used to solve first sum can solve the second one =)
so yeah try , why not !

Answer 10

ok what i know is using Fourier (but would be very messy)
but the other easy method is Zeta function
@ganeshie8 a question with zeta :)

Answer 11

You could probably use some variation of Euler's proof for \(\sum\dfrac{1}{n^2}\) converging to \(\dfrac{\pi^2}{6}\).

Where he used \(\dfrac{\sin x}{x}\), we (maybe) could get away with using \(\dfrac{\sin(x^3)}{x^3}\).

Answer 12

exactly :)

Answer 13

this is also euler approach he invented zeta

Answer 14

\[\zeta(2n) = \sum\limits_{i=1}^{\infty} \dfrac{1}{i^{2n}} = (-1)^{n+1} \frac{B_{2n} 2^{2n}}{2(2n)!} \pi^{2n}\]

plugin \(n = 3\) ?

Answer 15

Just curious, is there a way to write the infinite series as the limit of a finite series, and somehow translate that into a definite integral?

Answer 16

riemann sum @SithsAndGiggles ?

Answer 17

Yes I was wondering if there's a way to rewrite as follows
\[\sum_{n=1}^\infty \frac{1}{n^6}=\lim_{k\to\infty}\sum_{n=1}^kf(a+k\Delta x)\Delta x=\int_a^b f(x)\,dx\]
but I can't seem to make it work...

Answer 18

\[\int_0^1 x^{-k}\,dx = \lim\limits_{n\to \infty} \sum\limits_{i=1}^n (\frac{i}{n})^{-k} \cdot \frac{1}{n}= \lim\limits_{n\to \infty} \frac{1}{n^{-k+1}}\sum\limits_{i=1}^n i^{-k} \]

something like this ?

Answer 19

we know how to evaluate the left side definite integral
but extracting that sum on right hand side looks trickyk.. xD

Answer 20

that doesn't look converging, we might have to tweak the bounds hmm

Answer 21

yes exactly !

Answer 22

Here is an idea I had before but couldn't seem to take it very far, but I think I might have made some progress now, but I still need help. \[\Large \frac{\pi^2}{6} = \frac{1}{(1^1)^2}+\frac{1}{(2^1)^2}+ \frac{1}{(3^1)^2}+... \\ \Large \frac{\pi^4}{90} = \frac{1}{(1^2)^2}+\frac{1}{(2^2)^2}+ \frac{1}{(3^2)^2}+... \\ \Large \frac{\pi^6}{945} = \frac{1}{(1^3)^2}+\frac{1}{(2^3)^2}+ \frac{1}{(3^3)^2}+...\] But I was thinking, we can add up all the pi^2, pi^4, pi^6, etc terms to infinity and it would be equal to all of these geometric series if you look at them vertically going down the columns.\[\Large \sum_{k=1}^\infty \frac{1}{(n^k)^2}=\sum_{k=1}^\infty \left(\frac{1}{n^2} \right)^k=\frac{1}{n^2-1}\]

So I was thinking maybe there's some way to set these equal to each other and solve some thing like this and derive them from something like this (despite being divergent) \[\Large \sum_{n=1}^\infty \frac{1}{n^2-1} = \sum_{n=1}^\infty \frac{\pi^{2n}}{constant?}\] So here's as far as I can go, I don't know if there's a simple way to represent these constants in the denominator, but maybe we can reverse this into some kind of derivation just like how fourier series sort of work.

Answer 23

Actually on wikipedia it shows it as \[\Large \zeta (2n) = (-1)^{n+1} \frac{B_{2n}(2 \pi)^{2n}}{2(2n)!}\] and B_n are the Bernoulli numbers (which I really don't know much about sadly, so if anyone can explain them to me that would be great, so I can rewrite my equation as: \[\Large \sum_{n=1}^\infty \frac{1}{n^2-1} = \sum_{n=1}^\infty (-1)^{n+1} \frac{B_{2n}(2 \pi)^{2n}}{2(2n)!}\] Which is kind of interesting, since it looks like we have the left side can factor as a difference of two squares, I don't know if this gets us anywhere closer to solving Ikram's problem the long way or something lol.

Answer 24

There are two ways I can see how to change the stuff I threw up here to make it convergent and hopefully I can find something interesting.

Answer 25

Ok so instead of adding them all up divergently, let's just add zeta(2)-zeta(4)+zeta(6)-... etc and change the geometric series accordingly to have a negative sign in it! That will make it converge and change the expression to be: \[\Large \sum_{n=1}^\infty \frac{1}{n^2+1} = \sum_{n=1}^\infty \frac{B_{2n}(2 \pi)^{2n}}{2(2n)!}\] My first instinct is to associate the integral on the left with this integral: \[\Large \int\limits_0^\infty \frac{dx}{x^2+1}=\frac{\pi}{2}\] But then I also noticed that this summation on the left is really the shifted basel problem, which is strange, maybe there's an easy way to solve these geometrically afterall? I am sort of starting to believe that there is some geometric meaning by having pi in there. Maybe you find this interesting as well?

Answer 26

here is one way to generate bernouli numbers @Kainui
http://www.wolframalpha.com/input/?i=solve+w%2B4x+%2B+6y%2B4z%3D4%2C+2%5E4w%2B4*2%5E3x%2B6*2%5E2y%2B4*2z%3D36%2C+3%5E4w%2B4*3%5E3x%2B6*3%5E2y%2B4*3z%3D144%2C+4%5E4w%2B4*4%5E3x%2B6*4%5E2y%2B4*4z%3D400

Answer 27

\[\sum\limits_{k=1}^{n} k^m = \frac{1}{m+1} \sum \limits_{k=0}^m
\binom{m+1}{k} \color{Red}{B_k} n^{m+1-k}\]

we can generate all bernouli numbers by using that definition
http://en.wikipedia.org/wiki/Bernoulli_number

Answer 28

idk much about zeta function but i have seen bernouli numbers in number theory before
@Marki promised she will teach me zeta someday... fascinating stuff xD that integral reminds me of euler reflection formula :

\[\int\limits_0^{\infty} \dfrac{dx}{x^n+1} = \dfrac{\Gamma(1-\frac{1}{n}) \Gamma(\frac{1}{n})}{n} = \dfrac{\pi}{n\sin(\frac{\pi}{n})}\]

not sure if there will be some useful connection with the discrete sum :/

Answer 29

haha well u two said 2 imp. things
1- it really has something to do with solving Bessel equation
2-Euler derived it , no doubt ur gonna smell Euler :)
and @ganeshie8 i like this way in teaching step by step
i believe now ur be able to find zeta(2n) for solving series of this form :)
get ready for next step

Answer 30

@Marki Bessel equation, or Basel problem?

Answer 31

Basel, Bessel, Bernoulli... What's the difference? =P

Answer 32

i think problem thx for correcting me @SithsAndGiggles

Answer 33

Idk maybe there's some crazy connection to the Bessel equation too...

Answer 34

maybe , i'll see after exams

Answer 35

good luck wid exams @Marki

Answer 36

:)