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If An converges to 0&An =\=0 then An is:1)1/An is convergent(c)&bound...

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farahyasmin | Student, Undergraduate | (Level 1) Honors

Posted December 25, 2012 at 5:36 PM via web

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If An converges to 0&An =\=0 then An is:1)1/An is convergent(c)&bounded(b) 2)1/An is non-c&unb 3)1/An is c&unb 4)1/An is non-c&b

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted December 27, 2012 at 9:50 PM (Answer #1)

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For starters, if a sequence is unbounded, then it is not convergent.  If it converged to, say, a number L, then eventually all the entries of the sequence would fall within (L-1, L+1).  (This is like picking epsilon = 1.)  There would be finitely many numbers that DIDN'T fall in that range.  So, you've got a handful of numbers, possibly large, but only finitely many, so there's a biggest.  And the rest are sandwiched between L-1 and L+1, so they are bounded.  Thus the whole sequence is bounded.

What that means is, a sequence can never be unbounded and convergent. (which rules out #3)

So: if you can prove it converges, you automatically know it is bounded.  And if you can prove is it unbounded, you automatically know it is not convergent.

The sequence 1/An is neither convergent, nor bounded.


To see this, we first try an example.


Consider: {An} = 1, 1/2, 1/3, 1/4, 1/5, ...

Then, the sequence {1/An}  is  1,2,3,4,5,...

This sequence is unbounded.

 

 

 

Now, for the details:

If An converges to 0, then eventually all the numbers are within 1 of 0.  That is, |An|<1

But that means that |1/An|>1

So eventually, all of the numbers in 1/An are larger than 1 or less than -1

 

We can repeat this argument:

If An converges to 0, then eventually all the numbers are within 1/2 of 0.  That is, |An|<1/2

But that means that |1/An|>2

So eventually, all of the numbers in 1/An are larger than 2 or less than -2

 

If An converges to 0, then eventually all the numbers are within 1/10 of 0.  That is, |An|<1/10

But that means that |1/An|>10

So eventually, all of the numbers in 1/An are larger than 10 or less than -10

 

We can play this game for any number.

Thus, we can show that, eventually, all of the numbers in the sequence |1/An| are very large, as large as we would want.  Thus, the sequence is unbounded.

Thus, it can't be convergent.

 

 

 

PS:

In the example An=1, 1/2, 1/3, 1/4, ...

1/An = 1, 2, 3, 4, 5...

The sequence 1/An "diverges to infinity"

So, the sequence still has a limit, but the limit is infinity.

 

Even THAT doesn't always happen:

Consider:

An = 1, -1/2, 1/3, -1/4, 1/5, ...

Then 1/An = 1, -2, 3, -4, 5, -6

so the numbers aren't getting "closer and closer to infinity"

they are bouncing around while getting larger and larger in magnitude

 

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