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SOLUTIONOFHW3MINGFENGZHAOMarch 01, 20131. [10 Points] Compute the sup, inf, lim sup, lim inf, and all the limit points ofxn= (1)n+1n+2 sinn2for alln2N.Proof.Notice thatsinn2=8>>>>>>><>>>>>>>:1;ifn1 (mod 4)0;if 2jn1;ifn3 (mod 4):So we getxn=8>>>>>>><>>>>>>>:1 +1n;ifn1 (mod 4)1 +1n;if 2jn3 +1n;ifn3 (mod 4):Hence we havesupn1xn= 2infn1xn=3lim supn!1xn= 1lim infn!1xn=3All limit points offxngare 1 and3.1
2MINGFENG ZHAO2. [10 Points] IfEis a set andya point that is the limit of two sequences,fxngandfyngsuch thatxnis inEandynis an upper bound forE, prove thaty= supE. Is the converse true?Proof.Sinceynis an upper bound forEfor alln1, thenxyn;8x2E;8n1:Sinceyn!yasn!1, thenxy;8x2E:(1)That is,yis an upper bound ofE. On the other hand, sincexn!yasn!1, then for any >0,there exists someN2Nsuch that for allnN, we havejyxnj< :That is, we gety < xn+for allnN. Therefore, we know thaty= supE:Claim I: The converse is not true.LetE=R, then1= supE, but we can not nd anyyn2Rsuch thatyn!1, andynis upperbound forE.3. [10 Points] Prove lim sup (xn+yn)lim supxn+ lim supynif both limsups are nite, and givean example where equality does not hold.Proof.LetA= lim supn!1xnandB= lim supn!1yn, by the assumption, we know thatA;B2R. For any >0, by the de nition ofAandB, there exists someN2Nsuch that for allnN, we havexn< A+;andyn< B+:
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