This is not correct. You have to prove that two separate sequences are monotone, not that alternate "steps" in the original sequence are all in the same direction. What you need is for the sequences.

Reply 4

mathmari

Original post by BlueSam3

This is not correct. You have to prove that two separate sequences are monotone, not that alternate "steps" in the original sequence are all in the same direction. What you need is for the sequences.

How can I show this then??

Reply 5

TheIrrational

Original post by mathmari

a_{2k} - a_{2k-1} \leq 0 \Rightarrow 1+\frac{2}{1+a_{2k-1}}-a_{2k-1} \leq 0 \Rightarrow 1+a_{2k-1}+2-a_{2k-1}^2 \leq 0 \Rightarrow a_{2k-1}^2-a_{2k-1}-3 \geq 0 \Rightarrow a_{2k-1} \geq \frac{1+\sqrt{13}}{2}

,
since

a_{n} >0

for each n.

So how can I continue???

I think you've been given incorrect advice.

You wish to show that

a_{2k} - a_{2(k-1)}

is either less than or equal to 0 or greater than or equal to 0 for the first subsequence sequence and that

a_{2k-1} - a_{2(k-1) - 1}

is less than or equal to 0 or greater than or equal to 0 for the second sub sequence.

Also, what you've posted their you seem to have assumed what you were trying to prove and show it's correct, try starting with what you're given in the question and ending up with the identity :smile:

Reply 6

BlueSam3

Original post by mathmari

How can I show this then??

You need to show that

a_{2(k+1)} \geq a_{2k}

\forall k

, (or the opposite inequality), and the same for [tex]a_{2(k+1)+1} \geq a_{2k+1},

\forall k

, (or, again, the opposite inequality). There are two very standard ways of showing these things. Which looks easier in this case?

Reply 7

mathmari

And to show for example that

a_{2(k+1)+1} \geq a_{2k+1}

, I have to replace the relation given for

a_{n+1}

, right??

Reply 8

BlueSam3

Original post by mathmari

And to show for example that

a_{2(k+1)+1} \geq a_{2k+1}

, I have to replace the relation given for

a_{n+1}

, right??

Express

a_{2(k+1)+1}

in terms of

a_{2k+1}

Reply 9

DFranklin

Original post by BlueSam3

Express

a_{2(k+1)+1}

in terms of

a_{2k+1}

I wouldn't actually do that, although I'm partly speaking from the experience of having done very similar questions before.

Instead of worrying about

a_{2k}, a_{2k+1}

etc., simply express

a_{k+2}

in terms of

a_k

and then consider

a_{k+2} - a_k

.

Hint: At this point you may find it useful to represent a_k in terms of a_{k-2}.

Spoiler

Reply 10

mathmari

I found the difference

a_{k+2}-a_{k}

.To show that the subsequences are monotone, the difference must have contant sign, or not? How can I show this??

Reply 11

DFranklin

Post what you have so far.

Reply 12

mathmari

Original post by DFranklin

Post what you have so far.

I wrote the difference and I have this now:

a_{n+2}-a_{n}=\frac{3-a_{n}^2}{2+a_{n}}

, how can I show that the sign is constant?

Reply 13

DFranklin

What I meant for you to do is to find

a_{n+2}

in terms of

a_n

, and (seperately) find

a_n

in terms of

a_{n-2}

.

Use this to find

a_{n+2} - a_n

in terms of

a_n

and

a_{n-2}

. You should then be easily able to show that the sign of

a_{n+2}-a_n

is the same as

a_n - a_{n-2}

Reply 14

james22

The advice you have been given by DFranklin is the best way to do this, but if all else fails proving it by induction will almost always work in these sorts of cases.