Question

Prove 2^(1/4)*4^(1/8)*8^(1/16)*16^(1/32) ... ∞ = 2.

Alternative view : http://www.wolframalpha.com/input/?i=2%5E%281%2F4%29*4%5E%281%2F8%29*8%5E%281%2F16%29*16%5E%281%2F32%29+...+%E2%88%9E+%3D+2&dataset=&equal=Submit

Answer 1

another approach would be...

Answer 2

\[2^{1/4}\times4^{1/8}\times8^{1/16}\times...=2^{1/4}\times2^{2/8}\times2^{3/16}\times...\]

Answer 3

so what you are trying to prove is:\[\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+...=1\]

Answer 4

@Neeraj_OCW - do you have any ideas on how to prove this simpler identity?

Answer 5

@asnaseer No, i'm trying but not getting result, pls can u explain it using formula of sum to infinity i.e. S = a / (1 - r) as other things are out of my course for this time, like log.

Answer 6

It's not a geometric sequence, so I don't think S = a / (1 - r) will apply here.

Answer 7

I agree with @henpen

Answer 8

I think you may have to ponder this for a while to see if you can spot a nice pattern

Answer 9

give it a try - it is far more satisfying to solve something like this yourself. :)

Answer 10

It's essentially the thing that I deleted, proving that\[\sum_{n=1}^{ \infty}2^{-n-1-\log_2(n)}=1\]I know next to nothing about series manipulation, but there may be something in
\[\sum_{n=1}^{ \infty}2^{-n-1-\log_2(n)}= \sum_{n=1}^{ \infty}2^{-n}\]

Answer 11

Yes, it's not normal series, it's special series, the other questions like this are mixup of two series, but i'm not getting this.

Answer 12

Sorry, that should be +log

Answer 13

I've not studied log function yet.

Answer 14

I believe this is can be proved starting from this identity:\[\sum_{n=1}^{\infty}\frac{1}{2^n}=1\]

Answer 15

are you /allowed/ to base your proof or other existing series?

Answer 16

Yes.

Answer 17

and have you been told about the one I just posted above?

Answer 18

Yes.

Answer 19

Like this a = 1/2, r=1/2, then S=1

Answer 20

http://www.youtube.com/watch?v=mQTWzLpCcW0

Answer 21

ok, so I'll give you a hint and see if you can use it prove your identity.
1 = 1/2 + 1/4 + 1/8 + 1/16 + ...
therefore 1/2 = 1/4 + 1/8 + 1/16 + ...
and 1/4 = 1/8 + 1/16 + ...
etc

see if you can make use of this

Answer 22

Ok, trying.

Answer 23

No luck, pls help. I also tried splitting it into two series of G.P. but ...

Answer 24

See what you get if you add each column in here:

1/2 = 1/4 + 1/8 + 1/16 + ...
1/4 = 1/8 + 1/16 + ...
1/8 = 1/16 + ...
...
=======================

Answer 25

1. I got it, thanks.

Answer 26

yw :)