Suppose that for the sequence of real numbers {an}, lim sup (an) = c > 0 Prove t

Question

Suppose that for the sequence of real numbers {an}, lim sup (an) = c > 0

Prove that lim sup (an^2) = c^2.

For this question, I tried two ways:

1) Since c is the limsup of {an}, given e >0 and k>0, there is an N such that for all n>= N_1, an <= c + k, and for each N, there is an n_1 such that an_1 >= c - k. If we want an^2 <= (c+k)^2 = c^2 + k^2 +2ck <= c^2 + e, then we solve the quadratic inequality:

k^2 + 2ck - e <= 0. Since k>0, we have 0 < k <= (sqrt(c^2 + e)) - c

Similarly, if we want (an_1)^2 >= (c-k)^2 = c^2 + k^2 - 2ck >= c^2 -e, we need to solve the quadratic inequality:

k^2 - 2ck + e >= 0

The solution is

k>= c+sqrt(c^2 -e) OR k<= c - sqrt(c^2 -e); AND k>0. BUT HOW CAN WE MAKE SURE THAT c^2 - e >= 0 must hold???

Also, how to compare the magnitude of 0, c - sqrt(c^2 -e), c+sqrt(c^2 -e), and (sqrt(c^2 + e)) - c ???

The second way I tried is as follows:

2) Let S_N = sup{an | n>=N} for each N,

then for each fixed N and all n>=N,

an<=S_N

so since S_N is monotone decrasing with respect to N, and S_N converges to limsup(an) = c >0, we have S_N > 0 for each N.

BUT HOW CAN WE MAKE SURE THAT an >=0 FOR ALL n >=N AND FOR EACH N, so that we can assert:

(an)^2 <= (S_N)^2 ???

Now let's proceed with (an)^2 <= (S_N)^2, for all each fixed N and n >= N.

Then sup(an^2 | n >=N) <= (S_N)^2 for each fixed N

Take limit on both sides of the inequality.

Since for any sequence of real numbers {an}, {bn} such that {an} converges to a and {bn} converges to b, if an<=bn for each n, then we have a<b.

Thus, sup(an^2 | n >=N) <= (S_N)^2 for each N implies that

lim sup (an^2) <= lim N to infinity: (S_N)^2 = (lim N to infinity: S_N)^2 = (lim sup an)^2 = c^2

however, this only proves that  lim sup (an^2) <= c^2.

How to prove  lim sup (an^2) = c^2???

Thanks a lot for your help!

Explanation / Answer

Unless you restrict an in some way (for example, that it is positive), this statement is not true.

Consider the following sequence: 1, -2, 1, -4, 1, -6, 1, -8, ... (a2n+1 = 1, a2n = -2n for n an integer).

lim sup an = 1, but lim sup an^2 does not exist.

If an is confined to being positive, then there is a straightforward proof.

To show that lim sup an^2 = c^2, let delta = min(epsilon/3c,epsilon/3)

Then, as there exists an n for any delta such that max an < c + min(epsilon/3c,epsilon/3)

Then, max an^2 < (c + min(epsilon/3c,epsilon/3))^2

Consider 2 cases, c >= 1 and c < 1.

For c >= 1, max an^2 < (c + epsilon/3c)^2 = c^2 + 2/3 epsilon + epsilon^2/9c^2)   As epsilon can be arbitrarily small, then epsilon^2/9c^2 <= epsilon^2/9 < epsion/9 < epsilon/3, so c^2 + 2/3 epsilon + epsilon^2/9c^2 < c^2 + epsilon. But as there exists such a delta for any epsilon, as the lim sup of an exists, then lim sup an^2 <= c^2.

Working with (c - epsilon/3c) similarly shows that lim sup an^2 >= c^2, so lim sup an^2 = c^2.

For c < 1, the proof follows in a similar fashion. Now, min(c+epsilon/3c, c + epsilon/3) = c + epsilon/3, and (c+epsilon/3)^2 =

c^2 + 2/3 c epsilon + epsilon^2/9 < (as c <= 1 and epsilon is small) c^2 + 2/3 epsilon + epsilon^2/9 < c^2 + 2/3 epsilon + epsilon/9 = c^2 + 7/9 epsilon < c^2 + epsilon, so lim sup an^2 <= c^2. Working similarly with c - epsilon/3 shows that lim sup an^2 >= c^2, so lim sup an^2 = c^2 for c <= 1 as well.

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