N. Younger

So, I have to go. It's late. ;)

I have to discuss it tomorrow

yeah I also see that

do you also see that in the final formula for s?

I hope it's a nice result, since we see that s is proportional to ln(p')/ln(p)

OKidoki, thanks! I think I am ready :)

do you just squared the second last line to get the last one?

yeah I know okay, I just wanted to be sure that the coefficinet of t^2 is positive

I am just going through your first answer with the knowledge of all the proofs

I have understood all the proofs

nono I mean in your origin answer, for my origin problem

A part of the thesis yes

One last thing: How do you know that the factor a in your very first answer is positive?

Yeah I have understood it now

OKay

but why are we minimizing it instead maximizing?

But my intuition says that we than also want to maximize the remaing sum, but we are minimizing it

or?

Okay that means we showed that the whole remaining sum has a smaller absolute value than the dominating fifth term which has (+) sign

yeees that's what I wanted to hear ;)

to be negative we show that the remaining sum is in absolute value smaller than the fourth term -1/(6(p-1)^3)

or?

what means the remaining sum must be something negativ

we show that the first three terms are bigger than the whole infinite series

I mean look at the left bound

I just want to get the concept :)

but can you explain your cut version? why do we know that when we cut at (+) that everything behind is smaller?

Okay and for the left bound. The idea is to show that the absolute value of the (maximized) remaining infinite sum is also smaller than the absolute value of the fourth term -1/(6(p-1)^3)?

Can we say as the main idea for the proof, that we have to show that the absolute value of the (minimized) remaining infinite sum is smaller than the fifth term 1/(24*p^4) ?

OKay, can we say as the whole idea for the right bound

thus 1/120 < 1/24, which is true

LHS is maximal with p=2

at the end I have the inequality 1/120(p-1) < 1/24

+1?

its - 1 /(120(p-1)p^4)

or?

to get my remaining infinte sum \sum k=5 ^\infty (-1)/120*p^k

the geometric sum (-1) / 120(1-1/p) + the five beginning terms or?

so for the second bound I write

I think the idea is that we express our maximized remaining infinite sum by the geomteric series MINUS the beginning terms, which are to much or?

The equality is not right, I think

OKay yeah it'S difficult to explain. But Under your bold line

"120"-terms

so I can further minimize since "120"-are bigger, which we subtract

k! >= 120

for k>=5

and then k! > 120

yes I#ve done that

at the second I want to minimize the remaing infinite sum , I think

yeah I am okay with the first one