Example 941 Use the Root Test to determine whether the series X k 1 k k 1 k

# Example 941 use the root test to determine whether

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Example 9.41.Use the Root Test to determine whether the seriesXk=1k-k-1kconverges or diverges.Solution.Here we havelimk→∞krk-k-1k= limk→∞k-k-1= limk→∞k-k-11·k+k-1k+k-1!= limk→∞k-(k-1)k+k-1= limk→∞1k+k-1= 0.Sinceρ= 0<1, the Root Test concludes that the series converges.Theorem 9.42(Ratio Test).Given the seriesakfor whichak= 0for at most a finitenumber ofkvalues, letρ= limk→∞|ak+1/ak|.1.Ifρ[0,1), then|ak|converges.2.Ifρ(1,], thenakdiverges.Proof.Suppose that limk→∞|ak+1/ak|=ρfor someρ[0,1). Letδ >0 be sufficiently smallso thatr=ρ+δ <1. Then there exists some integerk0>0 such that 0≤ |ak+1/ak|< r <1for allkk0, so that|ak0+1|< r|ak0|,|ak0+2|< r|ak0+1| ≤r2|ak0|,|ak0+3|< r|ak0+2| ≤r3|ak0|,and in general,|ak0+k|< rk|ak0|for allk0. From this we obtain
230≤ |ak|<|ak0|rk-k0=|ak0|rk0rk(7)for allkk0.Since|ak0|/rk06= 0 and 0< r <1, Proposition 9.27 implies thatXk=0|ak0|rk0rkis a convergent geometric series, and thus so too isk=k0(|ak0|/rk0)rkby Proposition 9.23.Hence, in light of (7), the Direct Comparison Test implies that|ak|converges.Now suppose limk→∞|ak+1/ak|=ρfor someρ(1,]. Letδ >0 be sufficiently small sothatr=ρ-δ >1. Then there’s some integerk0>0 such that 1< r≤ |ak+1/ak|for allkk0,so that|ak0+1|> r|ak0|,|ak0+2|> r|ak0+1| ≥r2|ak0|,|ak0+3|> r|ak0+2| ≥r3|ak0|,and in general,|ak0+k|> rk|ak0|>0for allk0. Fromr >1 and|ak0|>0 we have limk→∞rk|ak0|=, so that limk→∞|ak0+k|=and we obtain limk→∞ak6= 0. Henceakdiverges by the Divergence Test.Remark.Given Proposition 9.39, whenever part (1) of the Ratio Test or Root Test is used toconclude that|ak|converges, it immediately follows thatakconverges as well.Some books present a version of the Ratio Test which maintains that the terms of a seriesakmust be positive, which is an unnecessarily restrictive hypothesis. All that is needed is torequire that the termsakbe nonzero for all but (at most) finitely many values of the indexk,as reflected in the statement of the Ratio Test above. If there are at most finitely many termsakequalling 0, then there will exist some integerk0such thatak6= 0 for allkk0, and thusthere need be no worries thatak/ak+1will be undefined due to division by 0.Example 9.43.Use the Ratio Test to determine whether the seriesXk=115k(k+ 1)42k+1converges or diverges.Solution.We evaluate limk→∞|ak+1/ak|forak= 15k/[(k+ 1)42k+1]:ρ= limk→∞ak+1ak= limk→∞ak+1·1ak= limk→∞15k+1(k+ 2)42k+3·(k+ 1)42k+115k
24= limk→∞15(k+ 1)(k+ 2)42= limk→∞15k+ 1516k+ 32=1516.Sinceρ= 15/16<1, the Ratio Test concludes that the series converges (absolutely).