For any natural number , prove that:
The question is from my sequences and series text.
Good night, fardeen gen. (Or good afternoon, evening, morning, I'm not sure what time of day it is in India)
First off, I believe that you have copied the equation wrong. It should be
Test at values of you should see that you need to flip the sign (unless I'm wrong)
Onto the solution
Solution
Step 1
Cross-multiply and then remove logs. The equation becomes
When we test , we see that equality holds.
Test and and we see that the inequality holds. Assuming that this pattern will hold true for all natural numbers greater than
We now wish to prove
Let
So we have
With a little algebraic manipulation we shall now prove
Assume that this is not true, i.e.
Because the coefficient of is positive, the curve has a max. turning point and because of our assumption it means that the entire curve is below the x-axis, i.e. it has no real roots.
For this equation to have no real roots the discriminant
Inserting values:
To determine the values for n under which our assumption holds, we use the quadratic formula
This value of is complex, however the question asks us to prove this statement of natural numbers.
Because our assumption holds true only when n is complex, we must reject our assumption due to question constrains.
Therefore, the statement must hold true for all natural numbers greater than .
Equality holds at .
Hence, our original statement
is true for all natural numbers.
End of solution
I'm not sure how valid, so I (and fardeen_gen) would appreciate it if a more experienced forum member critique this solution
Also, I would appreciate it if someone could critique my method of proof writing and give me tips on how to better my proof-writing skills.
Merci beaucoup in advance.
Hope this was useful
Its morning here And Good morning/evening/night to you too
I feel that the proof is in order(a bit unconventional though). I copied the question correctly. Printer's Devil for the book in all probability.
Any other ideas, anyone?
You might have copied the question correctly, but in problems like these, one can check the validity by substituting values. So assume you want to do it by induction. Lets start inducting from n=2.... What do you see?
@I-Think: Writing it as a quadratic expression is a no-no
Because in a quadratic expression the coefficients are constants. Here your co-efficient of p is depending on n and p itself is again dependent on 'n'. Thus its not a quadratic expression...
My most humble apologies for leading innocents astray. May a millstone be tied around my neck and I be thrown into the sea...
But before we get too carried away, I present as a sign of my remorse another solution. Take and critique, hopefully I won't have to walk the plank now.
Solution:
It has already been shown that equality holds when
So we shall prove that
Test , we have:
Assume true for :
If our assumption then :
From our assumption, by multiplying by we see that
Via algebra, we shall show that
Subtract from
This statement is positive for all natural numbers
Hence
And if
Then
And
End of solution
This should be better, isomorphism?
So hopefully I won't have to learn to swim with rocks tied around my neck...
Aaaah! More apologies necessary. Wherever there's , it's supposed to be , I'll edit it .
More editing to be done. I'll show them here. The final statements are supposed to be
Hence
And if
Then
Therefore
We have shown that is true when is true. And if is true, then by the Principle of Mathematical Induction, is true for all natural numbers
And our original statement
is true.
Am I safe now? I see sharks circling...