Thanks! Could it be, that you made a mistake in the last line? Since from $\frac{1}{(p-1)p} > \frac{1}{2(p-1)^2} $ I got $\frac{1}{2(p-1)} < \frac{1}{p}$.

you are correct, i will fix it , but the argument still holds.

I am not sure. With $p=2$ I got the contradiction $2<2$.

correct $p>2$ , but since $p=2$ have a solution, according to you, there is an explicit formula when $p=2$, so you can take $p>2$.

That's true! ;)

in the lower bound i break the exponential series at a positive term so i maximized it and in the upper bound i break the exponential series at a negative term so i minimized it, so if the proof is valid for the new expressions then it will hold for the whole exponential series.

Hey, are you there?

I am not really convinced about the last concern. I mean if you break the lower-bound exponential series, than the broken serie is smaller than the exponential series or?

no not really

okay, I am okay with all things but the lower and upper bound of the $\ln{p-1}{p}$

you can prove that 1-1/p+1/2p^2 -1/6p^3 < 1-1/p+1/2p^2-1/6p^3+1/24p^4+...

for example: for the left side (means lower bound), we can prove that the broken series is greater than the not broken series or?

if we can show this, than we can say, that the exp(-1/(p-1)) is a valid lower bound, or?

you need to prove that 1-1/(p-1)+1/2(p-1)^2+.... < 1-1/(p-1) +1/2(p-1)^2 < 1-1/p < 1-1/p+1/2p^2-1/6p^3 < 1-1/p+1/2p^2-1/6p^3+....

I prove the inside inequalities in the site

you are not sure about the outside inequalities containing infinite series

isn't that right ?

yes, that's exact the point. Your idea is clear for me. But how can we prove the outer inequalities?

O.K. i will be proving that on the site , since it requires the use of MathJax, O.K. ?

okay, thanks

as long as I can prrof the outer inequalites with formulas

okay , if the proof is troubling you can ask me in chat to clarify that

okay thanks, those are the left proofs I need, than I think the whole thing is clear

i prove one of the inequalities, and left the second for you to prove (i am too lazy , lol) , there proof is almost the same, but if you don't manage to prove it

i will modify the post with a proof for the second inequality as well.

okay thanks, I will go through now

the idea is that if you cut the infinite series at a (-) sign, then the finite series will be smaller than the infinite series while if you cut at a (+) sign, the the finite series will be bigger than the infinite series.

With cut at (-) sign, you mean, that the last term is negative or?

yes

observe that the infinite sum is alternating sum meaning it alter between (+) signs and (-) signs

okay

Why can you upper bound the remaining sum because of the (-1)^k

?

You just add terms instead of substracting and adding in mix, and that's why we can bound it from above?

Lets say we want to prove that 7>3 but i don't how to represent 3 mathematically in simple way but i know that 4> 3 and then i have to prove that 7>4 which is simpler to prove that 7>3 but 7>4>3

i make the argument more stronger if i need to prove that 7>1-2+3 but instead i proved that 7>1+2+3 then the proof is valid also for 7>1-2+3 , can you see why ?

Yes, okay. Or instead of alternating between -1 and 1 you are adding just positive terms, and therefor it is larger

correct ,so if i prove it for the large sum then its immediately true for the small sum

how do you get the line after "So our infinite sum is just"?

Since our geometric series is not starting from the beginning

i will answer that after the pray, indulge me a little bit

no problems

okay, i will update the post

since i need some mathematics writing

I am trying to prrof the oppposite bound

see the updated post starting from the bold line

and see if you are okay with the explenation

if you are fine with the first inequality proof, you could mimic the proof steps to prove the second inequality (aside from being lazy, if you proved the second inequality your self you will see it obviously in your eyes).

so, are you okay with the first inequality proof.

yeah I am okay with the first one

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

you are correct , for visual assume you want to prove that 1<7 ,but writing 7 mathematically is hard so instead i know that 6< 7 and i need to prove that 1<6

so we are trying to minimize the infinite sum

a step in the right direction is to treat (-1)^k as -1 because its always true that (-1)^k is -1 or 1 and both are bigger or equal to -1

yes I#ve done that

and then k! > 120

for k>=5

k! >= 120

right

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

"120"-terms

i did not understand what you are saying ?

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

The equality is not right, I think

its right, by using the geometric series sum and subtracting the terms that are not in the actual sum as a,ar,ar^2,ar^3

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?

correct if we maximized a term with (-) sign , its like as we are minimizing the whole thing for example : -5 < -4 but 5>4 so we maximized 4 to 5 and since its multiply by (-) sign we making sure its the minimized.

so for the second bound I write

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

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

or?

true, so you will have -1/(120(1-1/p))+1/120+1/120p+1/120p^2+1/120p^3+1/120p^4

can you simplify these terms by arithmetic ?

???

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

correct , +1

+1?

so you need to prove that 1-1/p+1/2p^2-1/6p^3 < 1-1/p+1/2p^2-1/6p^3+1/24p^4-1/120(p-1)p^4

no i am addicted to Math Exchange , +1 meaning good answer and i will vote up

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

LHS is maximal with p=2

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

correct

thus concluding the proof.

Q.E.D

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

not following what your are saying

please explain your last line

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) ?

yes that's the idea in general

with alternating sums

when we cut at (+) term, we know all the terms behind it are less than it , so we added small amount to the whole ,but when we cut at (-) term we know we subtracted small amount from the whole sum

what you are seeing as a pattern is true in general for alternating sum

also if you are satisfied with the proofs, please vote up in the site

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)?

yes

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

in simple word every term is bigger than all the infinite terms after it in absolute value, meaning we don't care about (+,-) sings.

isn't that what we just proved ?! , unless you are not satisfied with the method of proving

I just want to get the concept :)

I mean look at the left bound

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

what means the remaining sum must be something negativ

or?

yes

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

observe that the remaining sum when we cut at (+) sign starts with (-) sign and we proved the the first term is the dominant so the rest of the sum will be (-) smaller than 0 giving that when we cut at (+) its always bigger than the whole sum

yeees that's what I wanted to hear ;)

the same as when we cut at (-) sign the remain sum start with (+) term and since the first term is the dominant we know that it will add up to something bigger than 0 giving that when we cut at (-) its always smaller than the whole sum

happy to be help you

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

or?

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

yes

but why are we minimizing it instead maximizing?

here minimizing in the (-) sense ,look at things visually 7<12-3 and we need to minimize -3 meaning finding something smaller but in the (-) realm of integer -4 <-3 so showing that 7<12-4 will be effective for 7<12-3

going to pray, i will try to help you at my best afterwards

OKay

when proving inequalities, most of the times we are using intermediate inequalities which are easier to calculate, for when we try to prove expression1>expression2 we need to find expression3 and prove separately that expression1 > expression3 and expression3 > expression2 which will imply that expression1>expression2.

Yeah I have understood it now

is this problem your thesis for the university ?

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

A part of the thesis yes

i choose it to be 1/24 to match the sum in the inequality

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

I have understood all the proofs

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

1/p is positive

and 1/(p ln p)- 1/((p-1) p ln p) >0 multiply by p ln p to get that 1>1/(p-1) which is true when p>2

so a is positive

but be aware this a does not have something in common with the a parameter we used in the geometric series proof, it's just a shortage in the English letters to the all parameters we use in math.

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

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

yes since t=sqrt(s) and we solved for t but we want s

OKidoki, thanks! I think I am ready :)

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

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

i think s is proportional to p ln(p')/ln(p)

yeah I also see that

I have to discuss it tomorrow

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

good night