# Quick sequence questions

1. Nov 15, 2008

### squaremeplz

1. The problem statement, all variables and given/known data

I know that the statements: if lim sup|s_n|=5, then s_n is bounded, and if lim sup s_n = 0 then lim sup |s_n| are false but I can't think of counter examples? Can someone suggest one or two please. many thanks

2. consider the sequence s_n convergent. Define a new sequence t_n such that

t_n = s_n + (-1)^(n)*(s_n)

a. show that t_n diverges

well lim sup s_n = s

then t_n = s_n*(1 + (-1)^n)

and finally

1/s * lim sup t_n = lim (1+(-1)^n)

the last sequence diverges because no matter how big n gets

the set will be {0, 2/s, 0, 2/s, 0, 2/s} at some point.

b) show by any method that t_n has a convergent subsequence

if we look at the previous tail we see that the 2k terms of n give a convergent subsequence that is

{0,0,0,0..}

right?

thansk!

Last edited: Nov 15, 2008
2. Nov 16, 2008

### HallsofIvy

Staff Emeritus
Then you know wrong. If lim sup |s_n|= 5, then no subsequence can converge to anything larger than 5 or less than -5. That means that only a finite number of its terms can be larger than 5 or less than 5 and so the sequence is bounded by the largest of those and the smallest. If lim sup |s_n| is any finite number then the sequence s_n is bounded. I don't know what you mean by the second. "If lim sup s_n= 0 then lim sup |s_n|" is WHAT? Yes, it is false that the lim sup |s_n|= 0. s_n= 1/n if n is even, -1 if n is odd is a counter example.

Are you assuming that s is not 0? you are not told that.

No, it won't. It will "at some point" become a sequence in which odd terms are 0 and even terms converge to 2/s but not equal to what you have. And in any case, dirvergence of that sequence is not what you wanted to prove! Simpler is just to note that t_n itself is 0 for n odd, s_n for n even. In fact, the "theorem" as stated is not true. If s_n converges to 0, then t_n also converges to 0. You need the additional requirement that s_n NOT converge to 0 in order to argue that the sub-sequence {t_{2n}) converges to 2s while {t_{2n+1}) converges to 0. If 2s and 0 are different, {t_n} does not converge.

[/quote]
NO!!! t_n= s_n+ (-1)^n s_n is 0 for n odd: 2k+1, not 2k!

3. Nov 16, 2008

### squaremeplz

it sounds like you know this material very well. The instructions for problem 2 include the important fact that s_n be a convergent sequence with a limit not equal to 0. Also, thank you for pointing out the fact that t_n = 0 on odd terms.

I don't understand your example for s_n = 1/n on even and -1 on odd?

Xould you be more specific

the terms are {-1, 1/2, -1, 1/4, -1, ...}

lim sup s_n = 0?

Last edited: Nov 16, 2008