Inequality Proof Question: Help Needed (1 Viewer)

[unclepete667]

Hi,

I came across this inequality proof, which I could get parts 1 and 2 out, but not sure how to do part 3. Answers are very vague on what to do, and jump around alot. Any help would be appreciated, on how to approach part 3, or a worked solution that is more then 4 lines.

Thanks

 

[chilli 412]

isnt part 2 (e-1)? i might be tripping but isnt 1/1! + 1/2! + 1/3! +.... = e - 1 ? because the sum from n=0 to infinity of 1/n! = e
or is it in reference to the previous sum in part i

 

[unclepete667]

isnt part 2 (e-1)? i might be tripping but isnt 1/1! + 1/2! + 1/3! +.... = e - 1 ? because the sum from n=0 to infinity of 1/n! = e
or is it in reference to the previous sum in part i

Part 2 is in referencing the first part, and is really just a show LHS = RHS

 

[chilli 412]

Part 2 is in referencing the first part, and is really just a show LHS = RHS

oh word, any advice on learning this stuff early (preferably without buying a textbook yet) because i was thinking about doing 4u next year

 

[unclepete667]

Proof, in my view, is the worst topic in 4u. However, if you do want to look at it ahead of time, you can go through Eddie's Woo content or you can try Howard's Textbook, which is free (I'm not sure of what the BOS site thinks of the textbook, I like it, and find it ok). Proof is a lot of Trial and error for me, but if you want to look into stuff early, I'd go for proof or complex numbers.

 

[Lith_30]

Since this is a multi-part question, we probably need to use one of the previous parts.

From part ii we get that

now we need to show that the sum of is equal to the sum of n distinct divisors, so we should try to make n! the subject of the equation.

Simply reorganising the equation gives us a very messy answer which probably will not work.

An alternate path would be to multiply both sides by n!. This gives us n! on one side and a series of additions on the other, which is a good sign.

and this appears to be the solution, since we have n! in the numerator of every term except the last, all of them will be an integer cause the denominator is in between 2 and n!.

We know that this sequence will have n terms cause the first part of the equation (excluding that +1) will have exactly n-1 terms, hence including the +1 will give n terms.

We also know that this will give us divisors of n! because the denominator will cancel out with the numerator. ie which will give a factor of n!

Testing this out for 4!

note how there are exactly four terms

also notice how these are divisors of 24

as required

 

[unclepete667]

Since this is a multi-part question, we probably need to use one of the previous parts.

From part ii we get that

now we need to show that the sum of is equal to the sum of n distinct divisors, so we should try to make n! the subject of the equation.

Simply reorganising the equation gives us a very messy answer which probably will not work.

An alternate path would be to multiply both sides by n!. This gives us n! on one side and a series of additions on the other, which is a good sign.

and this appears to be the solution, since we have n! in the numerator of every term except the last, all of them will be an integer cause the denominator is in between 2 and n!.

We know that this sequence will have n terms cause the first part of the equation (excluding that +1) will have exactly n-1 terms, hence including the +1 will give n terms.

We also know that this will give us divisors of n! because the denominator will cancel out with the numerator. ie which will give a factor of n!

Testing this out for 4!

note how there are exactly four terms

also notice how these are divisors of 24

as required

Thanks alot. This makes a lot more sense then what they give as the suggested reponse.