Can Induction Prove All Terms in a Sequence Are Greater Than One?

[mtayab1994]

Homework Statement

(

Given: U0=2 and :

(U)n+1=(Un^2+Un)/(1+Un)

Homework Equations

prove that for every n in N : Un>1

The Attempt at a Solution

ok for n=0 i got Un+1=2 and 2>1 therefore for n=0 its true.

now we assume its true for U=0 and we want to show that Un+1>Un

((Un+1)^2+Un+1)/(1+(Un+1)^2) > Un

am i correct so far?

 

[mtayab1994]

Homework Statement

(

Given: U0=2 and :

(U)n+1=(Un^2+Un)/(1+Un)

Homework Equations

prove that for every n in N : Un>1

The Attempt at a Solution

ok for n=0 i got Un+1=2 and 2>1 therefore for n=0 its true.

now we assume its true for U=0 and we want to show that Un+1>Un

((Un+1)^2+Un+1)/(1+(Un+1)^2) > Un

am i correct so far?

sorry ((Un+1)^2+Un+1)/(1+(Un+1)^2)>1

 

[lanedance]

can't you just go
(U)n+1=(Un^2+Un)/(1+Un) = Un(Un+1)/(1+Un)=Un?

 

[vela]

Nope. With induction proofs, you have some statement P(n) that is either true or false for some value of n. In this case, your statement P(n) is "U_n is greater than 1." You first want to establish that P(n) is true for some specific value of n. Then you want to show that if P(k) is true, then P(k+1) is true.

Homework Statement

(

Given: U0=2 and :

(U)n+1=(Un^2+Un)/(1+Un)

Homework Equations

prove that for every n in N : Un>1

The Attempt at a Solution

ok for n=0 i got Un+1=2 and 2>1 therefore for n=0 its true.

For n=0, you're given U_n=2, which is obviously greater than 1. In other words, P(0) is true. You're not trying to show (at this step) that
$$U_{n+1} = \frac{U_n^2+U_n}{1+U_n} > 1$$

now we assume its true for U=0 and we want to show that Un+1>Un

Now you assume P(n) is true for n=k and show that this implies that it's true for n=k+1.

Your induction hypothesis is P(k) is true, that is, U_k>1, and you want to prove now that U_k+1>1. According to the recursion relation, you have
$$U_{k+1} = \frac{U_k^2+U_k}{1+U_k}$$You want to show the righthand side is greater than 1 because that means P(k+1) is also true.

((Un+1)^2+Un+1)/(1+(Un+1)^2) > Un

am i correct so far?

 

[mtayab1994]

Nope. With induction proofs, you have some statement P(n) that is either true or false for some value of n. In this case, your statement P(n) is "U_n is greater than 1." You first want to establish that P(n) is true for some specific value of n. Then you want to show that if P(k) is true, then P(k+1) is true.

For n=0, you're given U_n=2, which is obviously greater than 1. In other words, P(0) is true. You're not trying to show (at this step) that
$$U_{n+1} = \frac{U_n^2+U_n}{1+U_n} > 1$$
Now you assume P(n) is true for n=k and show that this implies that it's true for n=k+1.

Your induction hypothesis is P(k) is true, that is, U_k>1, and you want to prove now that U_k+1>1. According to the recursion relation, you have
$$U_{k+1} = \frac{U_k^2+U_k}{1+U_k}$$You want to show the righthand side is greater than 1 because that means P(k+1) is also true.

ok i did:

(Un^2+Un)/(1+Un^2)>1 then:

Un+1-1=(Un-1)/(1+Un^2) and since we know that Un>1 therefor Un-1>0 and 1+Un^2>0 so therefore Un+1>0

Is this correct?

 

[vela]

It would really help if you would typeset subscripts and superscripts as appropriate. It's really hard to understand what you've written. If you go in advanced more, there are tools to typeset your text. You could also use LaTeX.

Where did (U_n²+U_n)/(1+U_n²)>1 come from? Note that in your original post, U_n wasn't squared in the denominator.

 

[mtayab1994]

It would really help if you would typeset subscripts and superscripts as appropriate. It's really hard to understand what you've written. If you go in advanced more, there are tools to typeset your text. You could also use LaTeX.

Where did (U_n²+U_n)/(1+U_n²)>1 come from? Note that in your original post, U_n wasn't squared in the denominator.

sorry that was my fault Un is squared in the denominator. sorry. so that means that's true then.

 

[vela]

(Un^2+Un)/(1+Un^2)>1 then:

You don't start off with this. This is what you're trying to prove.

U_n+1-1=(U_n-1)/(1+U_n²)

This is fine because it only depends on the definition of U_n+1.

and since we know that U_n>1 therefore U_n-1>0 and 1+U_n²>0 so therefore U_n+1>0

Is this correct?

Yes, that's right.

For the actual proof you write down, you may want to start with something like
\begin{align*}U_{n+1} &= \frac{U_n^2+U_n}{1+U_n^2} \\
&= \frac{U_n^2+1+U_n-1}{1+U_n^2} \\
&= 1+\frac{U_n-1}{U_n^2+1}
\end{align*}And then explain as you did why that last term is positive. It makes it a bit clearer what your argument is, starting from a definition and ending with what you're trying to prove.