Sincesn≥0, we have-< sn-< s,∀n > N.Sinceis arbitrary,s >-impliess≥0.Corollary 0.3Supposesn→sandtn→twithsn≤tnfor alln. Thens≤t.Proof:Apply the theorem above totn-sn≥0.Theorem 0.41Let(sn)be a sequence withsn>0. Iflimn→∞sn+1sn=L <1,thensn→0.Proof:SinceL <1, we can choosec∈(L,1) and=c-L. Sincesn+1sn→L, this, thereexistsN >0 such thatsn+1sn-L <,∀n > N,i.e.,-<sn+1sn-L <,∀n > N,which impliessn+1sn< L+=c,∀n > N,1When (sn) decreases:s1≥s2≥s3≥...≥0, it is not necessary thatsn→0. For example,sn= 1 +1ndoes not converges to 0. However, this theorem says that if (sn) decreases “rapidly,”sn→0.62
and hencesn+1< csn,∀n > N.Then fixingn=N, for any positive integerk, we havesN+1< csN,sN+2< csN+1,sN+3< csN+2,......,sN+k< csN+k-1.ThensN+k< csN+k-1< c2sN+k-2< ... < ck-1sN+1< cksN.Since 0< c <1,ck→0 ask→0. By Theorem 4.2.4, we prove limsn= 0.[Example and remark]1. Note the difference betweensn+1sn<1andlimn→∞sn+1sn<1.Only the conditionsn+1sn<1 may not imply limsn= 0. For example, ifsn= 1-1n,which is decreasing, it satisfiessn+1sn<1butsn6→0. In fact, for this example, limsn+1sn= 1. Therefore, the condition limsn+1sn<1 is crucial.2. Letsn=1+n2n. Sincesn+1sn=1 +n+ 12n+1·2n1 +n=n+ 2n+ 1·12→12<1,by the theorem above, limsn= 0.3. Letsn=23n32n. Show limsn= 0. By applying Theorem above,sn+1sn=8n+19n+18n9n=89<1so that limn→∞sn+1/sn=89<1.63
4. Letsn=n22n. Prove limsn= 0.Sincesn+1sn=(n+ 1)22n+1·2nn2=12n+ 1n2=121 +1n2→12<1.Then applying the above theorem, limsn= 0.5. Letsn=n!nn. Sincesn+1sn=(n+ 1)!(n+ 1)n+1·nnn!=nn+ 1n→1e<1.By the theorem above, limsn= 0.Infinite LimitsDefinition1. A sequence (sn)diverges to+∞, denotedsn→+∞, if for each real numberM, thereis a positive integerNsuch thatsn> M∀n > N.
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- Limits, lim, Limit of a function, Limit of a sequence, Suppose sn
