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

## 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

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Math_QED
Homework Helper
2019 Award
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?

Math_QED
Homework Helper
2019 Award
Yes it is clear. Do you want me to re-post the question or understand what I'm asking?

sergey_le
thanks

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

Math_QED
Homework Helper
2019 Award
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?

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

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

Math_QED
Homework Helper
2019 Award
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.

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.

SammyS
Staff Emeritus
Homework Helper
Gold Member
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} ##
.

sergey_le