Sequence Formula

[Mathyes]

Analysis Class:

I have this problem here that have been struggle for hours and still could not figure out. Please give me some clues.

The Problem: Find a bounded sequence with three sub-sequences converging to three different numbers.

Solution: One possibility is 1,0,-1,1,0,-1,1,0,-1, .... BUT I CAN'T COME UP WITH THE FORMULA FOR THIS SEQUENCE (STUCK) !!!

\(\displaystyle {a_n} = \)

I have tried \(\displaystyle {a_{n = }}\sin (n(\frac{\pi }{2}))\), but It appears to work only for the first several terms.

 

[stapel]

Is there any restriction on the sequence? Or can you have three convergent subsequences because you have four (and three is less than four)? If so, then how about using your proposed expression, with the subsequences being created as 1, 0, -1, 0, 1, 0, -1, 0,...?

 

[Mathyes]

No, there is not any thing else saying beside as the book states,

The problem: Find a bounded sequence with three sub-sequences converging to three different numbers. The book then gives a hint as, "One possibility is 1, 0, -1, 1, 0, -1, 1, 0, -1, ...."

So I am assuming there must be a way to write such sequence. But if I still cannot come up with any thing close to that hint then I guess I have to try using the one from my expression \(\displaystyle {a_n} = \sin (n(\frac{\pi }{2}))\)

 

[Deleted member 4993]

Analysis Class:

I have this problem here that have been struggle for hours and still could not figure out. Please give me some clues.

The Problem: Find a bounded sequence with three sub-sequences converging to three different numbers.

Solution: One possibility is 1,0,-1,1,0,-1,1,0,-1, .... BUT I CAN'T COME UP WITH THE FORMULA FOR THIS SEQUENCE (STUCK) !!!

\(\displaystyle {a_n} = \)

I have tried \(\displaystyle {a_{n = }}\sin (n(\frac{\pi }{2}))\), but It appears to work only for the first several terms.

How about:

for n = 3*p - 2 → a_n = 1

for n = 3*p - 1 → a_n = 0

for n = 3*p .... → a_n = -1

for p = 1,2,3,................

 

[pka]

Analysis Class:
The Problem: Find a bounded sequence with three sub-sequences converging to three different numbers.
\(\displaystyle {a_n} = \)

This is not the example from your text, but has 0, 1, 2 as subsequence limits.
\(\displaystyle {a_n} = \dfrac{{{n^2} + 1}}{n} - 3\left\lfloor {\frac{n}{3}} \right\rfloor \)

 

[Mathyes]

Thank you for all your help,

Looking at PKA formula, " \(\displaystyle {a_n} = \frac{{{n^2} + 1}}{n} - 3\left\lfloor {\frac{n}{3}} \right\rfloor \), I have a question from \(\displaystyle 3\left\lfloor {\frac{n}{3}} \right\rfloor \). Do you mean 3 multiply \(\displaystyle \left\lfloor {\frac{n}{3}} \right\rfloor \) or you mean 3 and \(\displaystyle \left\lfloor {\frac{n}{3}} \right\rfloor \) (example \(\displaystyle 3\frac{1}{3}\)).

Now as I understand \(\displaystyle \left\lfloor {\frac{n}{3}} \right\rfloor \) is a floor function which will assign to \(\displaystyle \frac{n}{3}\) the largest interger that is less than or equal to \(\displaystyle \frac{n}{3}\).

I tried \(\displaystyle n = 2\) and plug into the formula and doesn't seem to come out as 0 or 1 or 2.

 

[pka]

Thank you for all your help,

Looking at PKA formula, " \(\displaystyle {a_n} = \frac{{{n^2} + 1}}{n} - 3\left\lfloor {\frac{n}{3}} \right\rfloor \), I have a question from \(\displaystyle 3\left\lfloor {\frac{n}{3}} \right\rfloor \). Do you mean 3 multiply \(\displaystyle \left\lfloor {\frac{n}{3}} \right\rfloor \) or you mean 3 and \(\displaystyle \left\lfloor {\frac{n}{3}} \right\rfloor \) (example \(\displaystyle 3\frac{1}{3}\)).

Now as I understand \(\displaystyle \left\lfloor {\frac{n}{3}} \right\rfloor \) is a floor function which will assign to \(\displaystyle \frac{n}{3}\) the largest interger that is less than or equal to \(\displaystyle \frac{n}{3}\).

I tried \(\displaystyle n = 2\) and plug into the formula and doesn't seem to come out as 0 or 1 or 2.

Each term of that sequence is distinct. None of then equals 0, 1, or 2.
However the subsequence \(\displaystyle a_{3n}=\dfrac{1}{3n}\to 0\), \(\displaystyle a_{3n-1}=1+\dfrac{1}{3n-1}\to 1\) and \(\displaystyle a_{3n-2}=2+\dfrac{1}{3n-2}\to 2\)

Look at this page.

 

[Mathyes]

Thank you for your explanation PKA,

 

[HallsofIvy]

Analysis Class:

I have this problem here that have been struggle for hours and still could not figure out. Please give me some clues.

The Problem: Find a bounded sequence with three sub-sequences converging to three different numbers.

Solution: One possibility is 1,0,-1,1,0,-1,1,0,-1, .... BUT I CAN'T COME UP WITH THE FORMULA FOR THIS SEQUENCE (STUCK) !!!

\(\displaystyle {a_n} = \)

I have tried \(\displaystyle {a_{n = }}\sin (n(\frac{\pi }{2}))\), but It appears to work only for the first several terms.

The problem says "Find a bounded sequence..." and your "1, 0, -1, 1, 0, -1, ... is exactly such a sequence. I see nothing that requires you to "write a formula".

But if you feel you must, "\(\displaystyle a_{3n}= 1\), \(\displaystyle a_{3n+1}= 0\), \(\displaystyle a_{3n+2}= -1\)" is perfectly good.

 

[Mathyes]

Thank you for your clarification,
Yes, I did not read the question carefully.