Simple inequality proof by induction

In summary, we showed that 1 + 1/4 + ... 1/n2 + 1/(n + 1)2 < 2 - 1/(n + 1)2, where 1/(n + 1) is a positive number.
  • #1
missavvy
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0
Hey guys... I just did this question... and I have the solution but I do not understand it still. If someone could just explain what happened.

Homework Statement



1 + 1/4 + ... 1/n2 < 2 - 1/n
For n>=2

Homework Equations





The Attempt at a Solution


So the base case:
n=2:
1/4 < 2- 1/2, 1/4 < 3/2 True

Suppose this holds for n:

1+ 1/4 +... 1/n2 + 1/(n+1)2 < 2 - 1/n + 1/(n+1)2

Here is where I don't understand what happens..
Since 1/n - 1/(n+1) = 1/n(n+1) > 1/(n+1)2 , we have 2 - 1/n + 1/(n+1)2 < 2 - 1/(n+1).
Then the inequality for n+1 follows.

So I'm confused with why we use 1/n - 1/(n+1).. where did it come from ?
Thanks!
 
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  • #2
Suppose this holds for n:

1+ 1/4 +... 1/n2 + 1/(n+1)2 < 2 - 1/n + 1/(n+1)2

I color coded that expression for you. We want to look at the expression on the left hand side and show that it is smaller than 2 - 1/(n + 1). The part that I colored in blue, is smaller than the part in blue on the right hand side, by the induction hypothesis. The part in red is just copied over. (It's just using that a + c < b + c, whenever a < b - for a, b, c > 0). Basically, we recognised the induction hypothesis and just literally substituted it.

So you want the entire thing to be less than 2 - 1/(n + 1). Therefore you need to show that 1/n + 1/(n+1)2 < 1/(n + 1).

The way I would show this is as follows:
[tex]\frac{1}{n} + \frac{1}{(n+1)^2} = \frac{1}{n+1} \left( \frac{n + 1}{n} + \frac{1}{n + 1} \right) = \frac{1}{n + 1} \left( 1 + \frac{1}{n} + \frac{1}{n + 1} \right) [/tex]
and since the last two terms in the brackets are positive, this is definitely smaller than
1/(n + 1).
 
  • #3
Thanks for the help.

I still don't understand the part to show 1/n + 1/(n+1)2 < 1/(n + 1). I don't understand where the teal coloured terms are coming from. Did you bring the 1/n over from the right side to the left? What happened to the 2 and why is there a 1/(n+1) still there on the right of the inequality if that is what happened?

Sorry if I'm over complicating things. Perhaps I'm just overanalyzing this..
 
  • #4
Do you understand what we are trying to prove?

We want to show that
1 + 1/4 + ... 1/n2 + 1/(n + 1)2 < 2 - 1/(n + 1), (*)
given that
1 + 1/4 + ... 1/n2 < 2 - 1/n. (**)

So we start on the left hand side, with the expression
1 + 1/4 + ... 1/n2 + 1/(n + 1)2
Let me put a pair of brackets there:
... = ( 1 + 1/4 + ... 1/n2) + 1/(n + 1)2
The part in brackets is now equal to the left hand side of equation (**). So if we add to both sides of (**) the positive number 1/(n + 1)2, we get
( 1 + 1/4 + ... 1/n2 ) + 1/(n + 1)2 < ( 2 - 1/n ) + 1/(n + 1)2.

That's the part where we used the induction hypothesis (**). The rest of the proof is just a sequence of more rewritings and inequalities to show that 2 - 1/n + 1/(n + 1)2 in turn is smaller than 2 - 1/(n + 1), as we set out to prove. There are probably a lot of ways to do this, the one I prefer (which I gave in my post) is to write
1/n - 1/(n + 1)2 = 1/(n + 1) * (1 + x) < 1/(n + 1) * 1 = 1/(n + 1),
where x is some positive number.
 

What is a simple inequality proof by induction?

A simple inequality proof by induction is a mathematical method used to prove that a statement is true for all natural numbers. It involves proving a base case and then showing that if the statement is true for one number, it is also true for the next number.

How is a simple inequality proof by induction different from other types of proofs?

A simple inequality proof by induction is unique in that it uses the principle of mathematical induction, which states that if a statement is true for a base case and it is also true for the next number, then it is true for all subsequent numbers. This differs from other types of proofs which may use different methods such as contradiction or direct proof.

What are the steps involved in a simple inequality proof by induction?

There are three main steps in a simple inequality proof by induction: 1) Proving the base case, which is showing that the statement is true for the first natural number. 2) Assuming the statement is true for some natural number, and using this assumption to prove that it is also true for the next natural number. 3) Concluding that the statement is true for all natural numbers by the principle of mathematical induction.

What are some common mistakes to avoid when using simple inequality proof by induction?

Some common mistakes to avoid when using simple inequality proof by induction include: not proving the base case, assuming the statement is true for all natural numbers instead of just the next number, and using circular reasoning. It is important to carefully follow the steps of induction and make sure all assumptions and conclusions are justified.

What are some examples of problems that can be solved using simple inequality proof by induction?

Simple inequality proof by induction can be applied to a wide range of mathematical problems, including proving inequalities, identities, and divisibility statements. Common examples include proving that the sum of the first n natural numbers is equal to n(n+1)/2, or that the Fibonacci sequence increases exponentially.

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