Question

how to simplify this
(1^-4 + 2^-4 + 3^-4 + 4^-4 + ...)/(1^-4 + 3^-4 + 5^-4 + 7^-4 + ...)

Answer 1

let
p = 1^-4 + 2^-4 + 3^-4 + 4^-4 + ...
q = 1^-4 + 3^-4 + 5^-4 + 7^-4 + ...
see your question is p/q.

now, we get the value of
p - q = 2^-4 + 4^-4 + 6^-4 + 8^-4 + 10^-4 + ... or
p - q = 2^-4(1^-4 + 2^-4 + 3^-4 + 4^-4 + ... )
p - q = 2^-4(p)
p - q = p/16
p - p/16 = q
(15/16) p = q
p/q = 16/15

so, it is 16/15

Answer 2

still confused :(

Answer 3

@RadEn 's argument is this
You can define your two variables, p and q, to be the numerator and the denominator of the expression. So you have
\[ p = 1^{-4} +2^{-4} + 3^{-4} +4^{-4} ..... = \sum_{n=1}^{\infty} n^{-4}\]
\[ q = 1^{-4} + 3^{-4} + 5^{-4} + 7^{-4}..... = \sum_{n=1}^\infty (2n-1)^{-4}\]
if you subtract those from each other, all of the odd bases cancel out
\[ \color{purple}p-q = ( \cancel{\color{purple}{1^{-4}} - 1^{-4}}) +\color{purple} {\textbf 2^{-4}} + (\cancel{\color{purple}{3^{-4}} - 3^{-4}} )+ \color{purple}{\textbf 4^{-4}} + (\cancel{\color{purple}{5^{-4}} - 5^{-4}}) + \color{purple}{\textbf 6^{-4}}..... \]
leaving you with
\[ p -q = 2^{-4}+4^{-4}+6^{-4}+8^{-4}...\]
from which you can factor out the 2^-4
\[ p - q = 2^{-4}(1^{-4}+2^{-4}+3^{-4}...) = 2^{-4}(p)\]
solve for q in terms of p
\[ q = p (1-\frac{1}{16})\]
Seeing then that
\[ \frac{p}{q} = \frac{16}{15} \]

Answer 4

thanks, @AllTehMaffs . i think that's make more sense now :)

Answer 5

It's a nice proof you found :)

Answer 6

but i cant explain it more, actually my englsih so bad. :)

Answer 7

* english

Answer 8

Your maths are awesome! So no worries ^^

Answer 9

moreover i cant using LATEX, if ican do like you can make be happy to see it
your writting is beautiful

Answer 10

:) thank you :)

Answer 11

thanks guys, that's make more understand now