Let a_n be sequence so that a_n+1-a_n>-1 and |a_n|>2 for all n.

[sergey_le]

Homework Statement

Let a_n be sequence so that a_n+1-a_n>-1 and |a_n|>2 for all n.
1.Prove that if there is a natural N such that a_N is positive, then a_n> 2 for all n
2. From paragraph 1 conclude, that almost all a_n are positive or that almost all a_n are negative.
3. Prove that if for every n is Happening a_n+1<\frac a_n a_1

Relevant Equations

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I need help only in section 3
I have some kind of solution but I'm not sure because it seems too short and too simple.
We showed in section 1 that an> 0 per n.
it Given that a_n + 1 <0 and a_n+1<\frac a_n a_1 In addition therefore a_1 <0 is warranted

 

[sergey_le]

I tried to use https://www.physicsforums.com/help/latexhelp/ .
But it didn't work, Maybe my question is clearer now?

 

[member 587159]

Here is what is my interpretation of your question. I'm not sure, the lack of parentheses makes the question ambiguous:

_________________________________

Let $a_n$ be sequence so that $a_{n+1}-a_n>-1$ and $|a_n|>2$ for all $n$.
1.Prove that if there is a natural $N$ such that $a_N$ is positive, then $a_n> 2$ for all n
2. From paragraph 1 conclude, that almost all $a_n$ are positive or that almost all $a_n$ are negative.
3. Prove that if for every $n$ is Happening $a_{n+1}<\frac {a_n}{a_1}$
Homework Equations:

I need help only in section 3.
I have some kind of solution but I'm not sure because it seems too short and too simple.
We showed in section $1$ that $a_n> 0$ per $n$.
Given that $a_{n + 1} <0$ and $a_{n+1}<\frac{a_n}{a_1}$. In addition therefore $a_1 <0$ is warranted.

_________________________________________

Is this your question? When you quote (click reply) my post, you will be able to see the Latex code to make it format like that. I suggest you take a loot at it so you learn what's going in. Essentially, what we are doing is placing double hashtags around math expressions. For example, instead of

a_n

write

##a_n##

This will format nicely as

$a_n$

Hope this is clear!

 

[sergey_le]

Here is what is my interpretation of your question. I'm not sure, the lack of parentheses makes the question ambiguous:

_________________________________

Let $a_n$ be sequence so that $a_{n+1}-a_n>-1$ and $|a_n|>2$ for all $n$.
1.Prove that if there is a natural $N$ such that $a_N$ is positive, then $a_n> 2$ for all n
2. From paragraph 1 conclude, that almost all $a_n$ are positive or that almost all $a_n$ are negative.
3. Prove that if for every $n$ is Happening $a_{n+1}<\frac {a_n}{a_1}$
Homework Equations:

I need help only in section 3.
I have some kind of solution but I'm not sure because it seems too short and too simple.
We showed in section $1$ that $a_n> 0$ per $n$.
Given that $a_{n + 1} <0$ and $a_{n+1}<\frac{a_n}{a_1}$. In addition therefore $a_1 <0$ is warranted.

_________________________________________

Is this your question? When you quote (click reply) my post, you will be able to see the Latex code to make it format like that. I suggest you take a loot at it so you learn what's going in. Essentially, what we are doing is placing double hashtags around math expressions. For example, instead of

a_n

write

##a_n##

This will format nicely as

$a_n$

Hope this is clear!

Yes it is clear. Do you want me to re-post the question or understand what I'm asking?

 

[member 587159]

Yes it is clear. Do you want me to re-post the question or understand what I'm asking?

I think I understand what you are asking but I have to go right now. I'll return to answer later if nobody else already has answered.

 

[sergey_le]

I think I understand what you are asking but I have to go right now. I'll return to answer later if nobody else already has answered.

thanks

 

[sergey_le]

I think I understand what you are asking but I have to go right now. I'll return to answer later if nobody else already has answered.

hiii
If you can help me with this problem I would be very happy

 

[member 587159]

hiii
If you can help me with this problem I would be very happy

Yes, sorry I had an exam yesterday and it took more time than I expected. I don't really understand what (3) wants me to prove? Can you explain?

You wrote:

3. Prove that if for every n is Happening a_n+1<\frac a_n a_1

I interpret this as "show that $a_{n+1} < \frac{a_n}{a_1}"$ for all $n$. Is that correct?

 

[sergey_le]

Yes, sorry I had an exam yesterday and it took more time than I expected. I don't really understand what (3) wants me to prove? Can you explain?

You wrote:

3. Prove that if for every n is Happening a_n+1<\frac a_n a_1

I interpret this as "show that $a_{n+1} < \frac{a_n}{a_1}"$ for all $n$. Is that correct?

I'm sorry I didn't write the problem to the end.
Yes I meant it #<$a_n+1$<$\frac a_n a_1$
Prove that if for every n is Happening $a_n+1$<$\frac a_n a_1$ then a1<0

 

[sergey_le]

Yes, sorry I had an exam yesterday and it took more time than I expected. I don't really understand what (3) wants me to prove? Can you explain?

You wrote:

3. Prove that if for every n is Happening a_n+1<\frac a_n a_1

I interpret this as "show that $a_{n+1} < \frac{a_n}{a_1}"$ for all $n$. Is that correct?

I meant what you wrote here.
If you don't understand what I wrote, tell me and I'll upload a photo of the inequality.
Because I can't post it with the forum tools

 

[member 587159]

I may be missing something but the statement looks wrong. Consider the sequence $(3n)_{n=1}^\infty= (3,6,9, \dots)$. Then clearly $a_{n+1}-a_n = 3(n+1)-3n = 3 > -1$ and $|a_n| = 3n > 2$ for $n \geq 1$, yet $a_{n+1} = 3(n+1) = 3n +3 < 3n/3 = n$ is definitely not true.

So probably I misinterpret the question after all.

 

[sergey_le]

I may be missing something but the statement looks wrong. Consider the sequence $(3n)_{n=1}^\infty= (3,6,9, \dots)$. Then clearly $a_{n+1}-a_n = 3(n+1)-3n = 3 > -1$ and $|a_n| = 3n > 2$ for $n \geq 1$, yet $a_{n+1} = 3(n+1) = 3n +3 < 3n/3 = n$ is definitely not true.

So probably I misinterpret the question after all.

Well thank you very much I will try to find a solution already.
Thanks also for your patience

 

[SammyS]

I'm sorry I didn't write the problem to the end.
Yes I meant it #<$a_n+1$<$\frac a_n a_1$
Prove that if for every n is Happening $a_n+1$<$\frac a_n a_1$ then a1<0

This simply a little help for you to use LaTeX.
You have the following for your second line:
Yes I meant it #<$a_n+1$<$\frac a_n a_1$
There are two pieces of LaTeX code in that sentence.
The first is: $a_n+1$
I assume you want that to be: $a_{n+1}$
To have more than one character in a subscript, enclose the string of characters within { and } , as follows:
$a_{n+1}$

Similarly for a fraction: To have more than one character in the numerator (top) and/or denominator (bottom), enclose each between { and }.

You had $\frac a_n a_1$

Change that to: $\frac {a_n} {a_1}$ giving you $\frac {a_n} {a_1}$ .

I often use \dfrac rather than \frac . That gives $\dfrac {a_n} {a_1}$ .

Finally, you can use the "<" symbol within the LaTeX code to make one large piece of code. (The resulting character looks better.)

$a_{n+1} < \dfrac {a_n} {a_1}$
.