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# Proof by induction Watch

1. Hi, I can do most basic proofs by the P.MI. easily but I'm struggling with this one:

Obviously, for the base case we just state
Then assume n=k is true and show that n=k+1 is true but I'm not sure how to do this?

Thanks for any help in advance.
2. (Original post by Math12345)
Assume for some .

Then for maximum (need larger numerator and smaller denominator) and for minimum (need smaller numerator and larger denominator).

So it is true for n+1 since .

Done.
You'd also have to explain/prove that it's an increasing sequence.

I would rewrite the fraction

Then it's clearer why .
3. (Original post by notnek)
You'd also have to explain/prove that it's an increasing sequence.

I would rewrite the fraction

Then it's clearer why .
Ah I see, thanks.
4. (Original post by notnek)
You'd also have to explain/prove that it's an increasing sequence.
Why is that?
5. (Original post by EricPiphany)
Why is that?
I think he's trying to say if i prove it is increasing then I can sub in the end points to show a_{n+1} is bounded between 1 and 2.

Otherwise 1.5 could be used for example etc.
6. (Original post by Math12345)
I think he's trying to say if i prove it is increasing then I can sub in the end points to show a_{n+1} is bounded between 1 and 2.

Otherwise 1.5 could be used for example etc.
7. (Original post by EricPiphany)
How so? By maximising and minimising the denominator (or vice versa)?

It's not really a justification since both the numerator and denominators are functions of ; so you need to optimise them jointly. (i.e: why is an = 1 the minimum and why is an = 2 the maximum? The answer is because it's an increasing sequence).

For example, between , it's not true that the minimum occurs at x=-1 and the maximum occurs at x=1 or at x=0 by "maximising the numerator and minimsing the numerator" those arguments don't work without justifying that it's an increasing sequence (which this one isn't)).
8. (Original post by Zacken)
How so? By maximising and minimising the denominator (or vice versa)?

It's not really a justification since both the numerator and denominators are functions of ; so you need to optimise them jointly. (i.e: why is an = 1 the minimum and why is an = 2 the maximum? The answer is because it's an increasing sequence).

For example, between , it's not true that the minimum occurs at x=-1 and the maximum occurs at x=1 or at x=0 by "maximising the numerator and minimsing the numerator" those arguments don't work without justifying that it's an increasing sequence (which this one isn't)).
Ok, I wasn't claiming that the argument could be generalised, only that it was justifiable in this case.
9. (Original post by EricPiphany)
Ok, I wasn't claiming that the argument could be generalised, only that it was justifiable in this case.
Only justifiable if it was an increasing sequence, which is precisely what notnek said, that he hadn't justified that it was an increasing sequence.
10. (Original post by Zacken)
Only justifiable if it was an increasing sequence, which is precisely what notnek said, that he hadn't justified that it was an increasing sequence.
I don't get what this has to do with whether the sequence is increasing or not.
11. (Original post by EricPiphany)
I don't get what this has to do with whether the sequence is increasing or not.
The bounds of the inequality are the point at which the extrema occur if it's an increasing sequence.
12. This is how I essentially understood the guy's post (I have obviously added a little rigour)

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