Limit of Bounding Products (Ext 2 Problem 16 style) (1 Viewer)

WeiWeiMan

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I've composed a problem so student can practise with problem 16s.
Let me know what do you think.

View attachment 51038
(i)
Let f(x) = ln(1+x) - x + x^2/2
f(0) =ln(1) - 0 + 0 =0
f’(x) = 1/(x+1) -1 + x
= x+1 + 1/(x+1) -2
> 2 -2 {by AM/GM for x>0, no equality since x =/= 0 => x+1 =/= 1/(x+1)
= 0
f’(x) > 0 for x > 0 and f(0) = 0
Thus, f(x) > 0 for x > 0
=> 0 < ln(1+x) -x + x^2/2
=> x - x^2/2 < ln(1+x) QED

(ii)
Let x = k/n^2 for k = 1,2,3,…, n

Via x-x^2/2 < ln(1+x) < x: [given]
1/n^2 - 1/2n^4 < ln(1+1/n^2) < 1/n^2
2/n^2 - 2^2/2n^4 < ln(1+2/n^2) < 2/n^2

n/n^2 - n^2/2n^2 < ln(1+n/n^2) < n/n^2

Adding these:
(1/n^2 + 2/n^2 + … + n/n^2) - (1/2n^4 + 2^2/2n^4 + … + n^2/2n^2) < ln[(1+1/n^2)(1+2/n^2)…(1+n/n^2)] < 1/n^2 + 2/n^2 + … + n/n^2

Factorising and simplifying a bit:

1/n^2 (1 + 2 + … + n) -1/2n^4 (1 + 2^2 + … + n^2) < ln(Pn) < 1/n^2 (1+2+… + n)

I am far too lazy to prove 1+2^2 + … + n^2 = n(n+1)(2n+1)/6 so I’ll just accept it as is

Using this alongside 1+2+…+n = n(n+1)/2 (Via AP):

n(n+1)/2n^2 - n(n+1)(2n+1)/12n^2 < ln(Pn) < n(n+1)/2n^2

Exponentiating all 3:
e^[n(n+1)/2n^2 - n(n+1)(2n+1)/12n^2] < Pn < e^[n(n+1)/2n^2]

I’ll do (iii) and (iv) later since I’m doing these on phone and it’s a genuine pain to type it out

My rough ideas are:

Sandwich/Squeeze, looks like sqrt(e)
Question literally tells you what to do
 
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vuhung

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Nice work Wei Wei.

You'll find some interesting findings working with part (iv), linking to part (i), Taylor expansion and the squeeze theorem.
 

WeiWeiMan

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is this even in the syllbus 💀
Almost certainly
(i), (ii) and (iii) can be very reasonably asked in 4U, however, (ii) is a bit annoying to do without the realistic ability to quote without proof that (1+2^2+3^2+...+n^2) = n(n+1)(2n+1)/6.

(iv) is nice to think about, however, not super HSC style I'm pretty sure. I think it's still a reasonable question since it tells you how you should arrive at the proof though.
 

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