vidal (mav3385) – HW14 – schultz – (53130)3converges. In the first casean=parenleftbigg6n+ 74nparenrightbiggnparenleftbigg23parenrightbiggn=parenleftbigg6n+ 76nparenrightbiggn,in which caselimn→ ∞an=e7/6negationslash= 0;by the Divergence Test, therefore the series∞summationdisplayn=1andiverges. On the other hand,bn=parenleftbigg6n+ 74nparenrightbiggnparenleftbigg−23parenrightbiggn= (−1)nparenleftbigg6n+ 76nparenrightbiggn= (−1)nan.But, as we have seen,limn→ ∞an=e7/6.Thusbnoscillates betweene7/6and−e7/6asn→ ∞; in particular,limn→ ∞bndoes not exist. Again by the Divergence Test,therefore, the series∞summationdisplayn=1bndiverges.Consequently, the given power se-ries does not converge atx=±2/3 and soitsinterval of convergence =parenleftbigg−23,23parenrightbigg.00510.0pointsDetermine the interval of convergence ofthe series1.parenleftbigg−54,−14parenrightbiggcorrect2.parenleftbigg−14,14parenrightbigg3.(−∞,∞)4.series converges only atx=−345.parenleftbigg14,54parenrightbiggExplanation:The given series has the form∞summationdisplayk=0ck(x−a)kwithck=k32k,a=−34.But then,limk→ ∞ck+1ck=limk→ ∞2parenleftBigk+ 1kparenrightBig3= 2.By the Ratio Test, the series thus(i) converges when|x−a|<12, and(ii) diverges when|x−a|>12.Now at the pointsx−a=12andx−a=−12the series reduces to∞summationdisplayk=0k3,∞summationdisplayk=0(−1)kk3respectively. But in both cases these series di-verge by the Divergence Test. Consequently,the interval of convergence of the given seriesisparenleftbigga−12, a+12.

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