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"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=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!

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  • $\begingroup$ There is another question which I'm confused about: let {an} be a sequence of real numbers and suppose that limsup(an) is finite. Let {cn} be another sequence and suppose that {cn} converges to c. Prove that if c>=0, then limsup(cn * an) = c*limsup(an) $\endgroup$ – user118464 Dec 31 '13 at 15:53
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    $\begingroup$ Can you try to use $\LaTeX$? $\endgroup$ – Pratyush Sarkar Dec 31 '13 at 15:57
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I believe the problem as stated is incorrect.

For example, if we let $a_n=\begin{cases}-2, & \text{if n is odd} \\ 1+\frac{1}{n}, & \text{if n is even} \end{cases}$

then $\limsup a_n =1$, but $\limsup \;(a_{n})^2=4$.

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  • $\begingroup$ Thanks for your answer. But the problem states that lim sup (an) = c > 0. Your counter example has limsup(an) < 0... $\endgroup$ – user118464 Jan 1 '14 at 2:32
  • $\begingroup$ I think this counterexample is all right, since $\limsup a_n=1>0$ $\endgroup$ – user84413 Jan 1 '14 at 18:44
  • $\begingroup$ Thanks! But in my problem, it says "limsup{an}=c > 0" In this case, the statement would be correct, right? $\endgroup$ – user118464 Jan 5 '14 at 2:29
  • $\begingroup$ No, I don't think it is correct: $\limsup a_n=1>0$ in this case, but $\limsup (a_n)^2\ne1^2$. If the problem specified that $a_n\ge0$ for all n, then I think it would be correct. $\endgroup$ – user84413 Jan 5 '14 at 16:28

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