How would you calculate the sum to infinity (or just to n terms)of the series:

(ln 1)/2 + (ln 2)/4 + (ln 3)/8 + ..+ (ln(n))/2ⁿ

I am trying and i am trying!!

love arun

me too!! I think it converges around the 0.5078 mark, but i want to know why, and if it is either transcendental or can be expressed in exact form.

Well it is quite easy to understand that answer is between 0 and 1.

Applying little of my common sense i see that,
The terms are so arranged that the final answer is approximately 0.5 just because as n®¥ logn/2ⁿ tends towards zero.
However, i could not come up with any good proof or anything.

One more thing I have heard this word "transcendental" time and again after i have joined nrich but i never really understood its meaning.

Can you please tell me?

love arun

A transcendental number is a number that isn't the root of any polynomial equation with integer coefficients. Both e and p are transcendental, though I don't know how to prove this.
Olof

Another more elegant way of putting is it by saying transcendental numbers are numbers which 'transcend' our number system. Our decimal system is one particular one, we count 1, 2, 3 and so on. You could have another system say that counted in p: 1p, 2p, 3p ... and so on. They may never have heard of this number '1'. 1 transcends their number system, they can calculate in terms of p, it would be approximately 0.3183 ×p, but they could never find an exact value for it. They would have to give it a special symbol, like we do p.

It is almost as if p and e are from a higher 'plane' of numbers!

Ya i got it.

These numbers are something which are alien to the world & customs of algebraic numbers.am i right?

love arun

correct:)

Are proofs that p and e are transcendental very difficult?

Andrew - I was thinking about that very thing just the other day. Wouldn't their number system be identical to ours, but just with different symbols? As in pÞ 1, 2pÞ 2, etc. I'm not sure, but isn't there really only one 'counting system' (I can't think of the right phrase) when talking about real numbers? With our base, 1, we have that 2 = (1+1)×1. If we were using ps as our base, then we'd have 2p = (p+ p)×p, which is clearly not true, unless p = 1, i.e. the same base as ours? It's all very confusing!

Regards,
Olof

OLOF,
I can quite understand your problem.
I was just thinking,
What would be your problem converted into if we take it as,
2p = (1+1)×p

or

2p = (p+p)×1
arguments anybody!!!

love arun

The proofs that e and p are transcendental are indeed quite hard. They both follow from Lindemann's theorem which states that if n is natural and a₁, ..., a_n are algebraic numbers and b₁, ..., b_n are distinct algebraic numbers then:

a₁exp(b₁) + ¼+ a_nexp(b_n)

is non-zero.

When I get time over the next few days I'll write out the proof of Lindemann's theorem, perhaps in a new thread to stop it cluttering this thread - the proof is pretty complicated!

Anyway, how does this help with e and p?

Well if e is algebraic then if you set n = 2, a₁ = e, a₂ = 1, b₁ = -1, b₂ = 0 then since a₁, a₂, b₁, b₂ are all algebraic (with b₁ ¹ b₂) we deduce by Lindemann's theorem that the following is non-zero:

e ×exp(-1) + (-1) ×exp(0)

This is clearly nonsense, so we are forced to conclude that e is transcendental.

To prove that p is transcendental, assume it is algebraic. It easily follows that ip is algebraic (can you see why?) Hence using Lindemann's theorem, with n = 2, a₁ = 1, a₂ = 1, b₁ = ip, b₂ = 0 we deduce that the following is non-zero:

exp(pi) + 1

As we all know, the expression does indeed equal zero, so we have our contradiction proving that p is transcendental.

Take your own time michael.
I'll be waiting.

Boy,this is getting interesting!!
i am getting to learn something new everyday!!

love arun