Series is divergent 3 absolutely convergent

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2.series is divergent3.absolutely convergentExplanation:The given series can be written as an alter-nating seriessummationdisplayn=1(1)nanwithan= sinparenleftbigg15nparenrightbigg>0.To test for absolute convergence,i.e., conver-gence of the seriessummationdisplayn=1an,we use the Limit Comparison test withan= sinparenleftbigg15nparenrightbigg,bn=15n.
Version 1063 – EXAM 3 – schultz – (56190)7For thensummationdisplayanandsummationdisplaybnare series withpositive terms andlimn→ ∞anbn=limn→ ∞sinparenleftbigg15nparenrightbigg15n=limθ0sin(θ)θ= 1>0.Thus the given seriessummationdisplayn=1sinparenleftbigg15nparenrightbiggconverges if and only if the seriessummationdisplayn=115nconverges. But by thep-series test withp= 1(or because the harmonic series is divergent),this last series is divergent.Thus the givenseries is not absolutely convergent.On the other hand, sincesinparenleftbigg15nparenrightbigg>sinparenleftbigg15(n+ 1)parenrightbiggwhilelimn→ ∞sinparenleftbigg15nparenrightbigg= 0,the Alternating Series Test applies, showingthat the seriessummationdisplayn=1(1)n1sinparenleftbigg15nparenrightbiggis convergent. Consequently, the given seriesisconditionally convergent.012B.summationdisplayk=14k2 + 3kare convergent.1.both of them2.neither of themcorrect3.A only4.B onlyExplanation:A. Divergent:use Limit Comparison Testandp-series Test withp= 1.B. Divergent:use Limit Comparison Testand Geometric Series.01310.0pointsDetermine whether the series4163+64925627+· · ·is convergent or divergent, and if convergent,find its sum.1.series is divergentcorrect2.convergent with sum = 33.convergent with sum =974.convergent with sum =875.convergent with sum = 4
10.0pointsDetermine which, if any, of the seriesA.summationdisplayn=23nnn2
Explanation:The infinite series4163+64925627+· · ·=summationdisplayn=1a rn1
Version 1063 – EXAM 3 – schultz – (56190)8is an infinite geometric series witha= 4,r=43.But an infinite geometric seriessummationdisplayn=1a rn1(i) converges when|r|<1 and hassum =a1rwhile it(ii) diverges when|r| ≥1.Consequently, the givenseries is divergent.01410.0pointsDetermine whether the seriesI.summationdisplayn=1nn32n+1,

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