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Real Analysis IIHomework 6H´ector Guillermo Cu´ellar R´ıosMarch 30, 200619.2(b) The set of subsequential limits of a bounded sequence is always non-empty.True by Corollary 19.12.(c) (sn) converges tosiff lim infsn= lim supsn=sTrue by third paragraph in Definition 19.9 and exercise 19.9.(d) Let (sn) be a bounded sequence and letm= lim supsn. Then foreveryε >0 there are infinitely many terms in the sequence greaterthanm-εTrue by Theorem 19.11(b)19.3 For each sequence, find the setSof subsequential limits, the limit superior,and the limit inferior.(a)sn= (-1)nS={-1,1}lim supsn= 1lim infsn=-1(c)un=n2[-1 + (-1)n]S={-∞,0}lim supun= 0lim infun=-∞19.4 Let (sn) be a bounded sequence and suppose that lim infsn= lim supsn=s. Prove that (sn) is convergent and that limsn=s.(a)wn=(-1)nnS={0}lim supwn= 0lim infwn= 0(d)zn= (-n)nS={-∞,+∞}lim supzn= +∞lim infzn=-∞19.5 Use exercise 18.14 to find the limit of each sequence.1
(a)sn=(1 +12n)2nsn=(1 +12n)2n→esince it is a subsequence of the sequencesn=(1 +1n)n(b)sn=(1 +1n)2n(1 +1n)2n=(1 +1n)n(1 +1n)nsince each sequence is converging toe, then by Theorem 17.1(c)(1 +1n)2n→e2(c)sn=nn+1nnn+1n=1(n+1n)n=1(1+1n)n→1e=e-1(e)sn=(1 +12n)n(1 +12n)n=(1 +12n)2n2=(
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Fall '08
Akhmedov,A
Limits, Limit of a sequence, Sn, subsequence, lim sup sn, Snk