Question

Evaluate:
(1 + 1/3^5 + 1/5^5 + 1/7^5 + ...)/(1 + 1/2^5 + 1/3^5 + 1/4^5 + ...)

Answer 1

31/32

Answer 2

What a strange question. This is beyond me!

Answer 3

1+1/2^5 + 1/3^5 + 1/4^5 +...
(1/2^5 + 1/4^5 + 1/6^5+...) + (1+1/3^5 + 1/5^5+...)
1/2^5( 1+1/2^5 + 1/3^5+...)+(1+1/3^5+1/5^5+...)

Answer 4

So,
1+1/2^5 + 1/3^5 + 1/4^5 +...=1/2^5( 1+1/2^5 + 1/3^5+...)+(1+1/3^5+1/5^5+...)
(1-1/2^5) (1+1/2^5+1/3^5+1/4^5+..) = (1+1/3^5+1/5^5+...)

Answer 5

Now,
(1 + 1/3^5 + 1/5^5 + 1/7^5 + ...)/(1 + 1/2^5 + 1/3^5 + 1/4^5 + ...)
{(1-1/2^5) (1+1/2^5+1/3^5+1/4^5+..) }/ (1 + 1/2^5 + 1/3^5 + 1/4^5 + ...)
(1-1/2^5)/1
(2^5-1)/2^5
31/32

Answer 6

Oh I didn't see the division part.. I was trying to compute something completely different.

Answer 7

thanks @sauravshakya !
u made it seem not too difficult....
though wolf told me \(\sum \limits_1^\infty (1/n^5)=\zeta(5)\)
so i was wondering what's \(\zeta(x)\).....but with our method, it isn't required anymore....

Answer 8

Those are Reimann zeta functions that are quite peculiar.. one of them relates prime numbers to pi.