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=0(-1)n√n, which converges by Leibnitz’ theorem. Forx= 2, we get∑∞n=01√nwhich diverges (see this either by the integral test or comparison with the harmonic series).Combining this together, we get that the interval of convergence is [0,2), and the radius ofconvergence is 1.Example 51.Find the interval of convergence and radius of convergence of∑∞n=1n√n(2x+ 5)nSolution.First, we have to express the given function as a power series:∞Xn=1n√n(2x+ 5)n=∞Xn=1n√n(2n)x+52n(in particular, note thata=-52, not-5). Using then-th root test, we see(1)nand ifx+52=-12,thenx=-3 and the corresponding series is∑∞n=1n√n(-1)n. Neither of these series hasan→0, sothey both diverge. This means that the interval of convergence is (-3,-2).7.2.Term-By-Term Calculus.If∑cn(x-a)nconverges on some positive radius of convergence(i.e. it doesn’t diverge for everyx6=a), thenf(x) =∑cn(x-a)nis a continuous function (not todifficult to show), and we can reduce taking derivatives and integrals offto taking derivatives andintegrals of the terms (more difficult to show):7.2.1.Term-By-Term Differentiation.Theorem 7.3.Iff(x) =∑∞n=0cn(x-a)nconverges on|x-a|< R, then:(1)fhas derivatives of arbitrary order on(a-R, a+R)(i.e.f0(x), f00(x), f000(x), . . .exist)(2)The derivatives are given by:f0(x) =∞Xn=1ncn(x-a)n-1, f00(x) =∞Xn=2(n)(n-1)(x-a)n-2, . . .on(a-R, a+R).We can use this to express some power series in other ways:Example 52.Recall thatf(x) =∑∞n=0xnconverges for-1< x <1. The above theorem tells usthatf0(x) =∞Xn=1nxn-1andf00(x) =∞Xn=2n(n-1)xn-2On the other hand, we know thatf(x) =11-x, sof0(x) =1(1-x)2andf00(x) =2(1-x)3. This gives usformulae:∞Xn=1nxn-1=1(1-x)2,and∞Xn=2n(n-1)xn-2=2(1-x)3It is important to note here thatcndon’t involvexin any way (otherwise the series would notbe a power series). It is not true that if we take any series∑fn(x) of functions ofxthat we candifferentiate term-by-term as above:Example 53.The series∑∞n=1sin(n!x)n2converges absolutely for allx, but if we differentiate term-by-term we get∑∞n=1n! cos(n!x)n2which diverges for allx.
MATH 214: SEQUENCES AND SERIES137.2.2.Term-By-Term Integration.Theorem 7.4.Givenf(x) =∑∞n=0cn(x-a)nthat converges on|x-a|< R, the functionF(x) =∑∞n=0cn(x-a)n+1n+1converges on|x-a|< RandZf(x) dx=F(x) +CWe can use this to identify series with other functions as well:Example 54.Identifyf(x) =∑∞n=0-(1-t)n+1n+1on0< x <2.Solution.First, differentiate the given series term-by-term to getf0(x) =∞Xn=0(1-t)n=11-(1-t)=1tThis means thatf(x) =Rf0(x)dx+C= ln(x)+C. Settingx= 1 we getC= 0, i.e.f(x) = ln(x).