Proving the Convergence of (y_n) Given a Properly Divergent Sequence (x_n)

[ILikePizza]

Homework Statement

Suppose that (x_n) is a properly divergent sequence, and suppose that (x_n) is unbounded above. Suppose that there exists a sequence (y_n) such that limit (x_n * y_n) exists. Prove that (y_n) ===> 0.

Homework Equations

(x_n) ===> 0 <====> (1/x_n) ===> 0

The Attempt at a Solution

One can say with certanty that (y_n) must be bounded, as if it weren't, for all K in Naturals, there exists a b_1 in (x_n) > |K| and b_2 > |K|, and there product is unbounded.

If (y_n) is bounded, and does not converge to 0, then... what?

That's where I'm stuck. How do I finish this?

Thanks.

 

[Dick]

If y_n does not converge to zero then there is an e>0 such that for all N there is an n>N such that |y_n|>e. If x_n is unbounded, what does this tell about y_n*x_n?

 

[boombaby]

When I was trying to prove it directly (for fun), I met some problems.
What is a *properly* divergent sequence?( I cannot find its definition in books)
If x_n is defined as follows:
x_n = 0 when n is even, x_n = n when n is odd
is it of such kind?
If so, define y_n as:
y_n = 1 when n is even, y_n = 0 when n is odd

does this gives x_n*y_n = 0, as a counter??

 

[HallsofIvy]

When I was trying to prove it directly (for fun), I met some problems.
What is a *properly* divergent sequence?( I cannot find its definition in books)
If x_n is defined as follows:
x_n = 0 when n is even, x_n = n when n is odd
is it of such kind?
If so, define y_n as:
y_n = 1 when n is even, y_n = 0 when n is odd

does this gives x_n*y_n = 0, as a counter??

The crucial part of the problem is "suppose that (x_n) is unbounded above". Your example does not satisfy that.

 

[Office_Shredder]

A properly divergent series is one such that

(mathematics) A series whose partial sums become either arbitrarily large or arbitrarily small (algebraically).

So turning each partial sum into an element of the sequence, I believe a properly divergent sequence is one in which for all M there exists k s.t. j>k => |x_j|>M

 

[boombaby]

A properly divergent series is one such that

(mathematics) A series whose partial sums become either arbitrarily large or arbitrarily small (algebraically).

So turning each partial sum into an element of the sequence, I believe a properly divergent sequence is one in which for all M there exists k s.t. j>k => |x_j|>M

Thanks..I forgot to search the web... That would make sense. So a direct proof is also not hard.

BTW, to HallsofIvy, my x_n do satisfy the unboundedness, IMO.