1549901369-Elements_of_Real_Analysis__Denlinger_

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70 Chapter 2 • Sequences

We have already had occasion to make use of this fact. An "eventually
constant" sequence (Definition 2.2.4) is one that has a constant tail; in Theorem
2.2.5 we saw that the constant tail determines its convergence. Again, in stating
and proving Theorem 2.2.13 (g), we ignore all terms of {xn} that come before
the n 1 th one (that is, all terms where vx;;, might not exist) and deal only with
the n 1 -tail. Now we know why we can do that.


EXERCISE SET 2.2

./ l. Prove Theorem 2.2.l.



  1. Show by example that the converse of Theorem 2.2.l(c) is not true.


...--3. Prove Theorem 2.2.3.

/



  1. Prove Theorem 2.2.5.

  2. Give an example of a bounded sequence that does not converge.

  3. Prove that if {an} is a bounded sequence (not necessarily convergent)
    and {bn} converges to 0, then anbn -; 0. Use this result to prove that
    lim sin n = lim cos n = 0.
    n-+(X) n n-+(X) n

  4. Give an example of sequences {an} and {bn} for which bn -; 0, but
    anbn ~ 0.

  5. Prove Parts (a) and (b) of Theorem 2.2.12 as easy consequences of (c).


/9. Prove Theorem 2.2.13 (a), Case 2.

10.
/

Give an example of sequences {an} and {bn} for which {an +bn} converges
but one or both of {an} and {bn} does (do) not. In such a case, we cannot
say lim (xn + Yn) = lim Xn+ lim Yn· Does this contradict Theorem
n-+oo n--+cx:> n-+oo
2.2.13 (b)? Explain.


  1. Prove Part (c) of Theorem 2.2.13 as an easy consequence of (a) and (b).

  2. Give an example of sequences {an} and {bn} for which {anbn} converges
    but one or both of {an} and {bn} does (do) not. In such a case, we cannot
    say n-+oo lim (XnYn) = n-+cx:> lim Xn · n-+oo lim Yn· Does this contradict Theorem 2.2.13
    (d)? Explain.

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