Series0203.pdf - 3 Series 1 X You may want to think of a...

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3SeriesYou may want to think of a series of real numbers,anas an “infinite sum”, but thatnotion is too vague and may lead to confusion and error. We define an infinite series as alimitof finite sums, and so it is just another example of a sequence!Definition 3.1.Given a sequence(an)n2NinR, we define the series1Xn=1anas follows: foreachk2N, define thekthpartial sumsk=kXn=1an.We say the series1Xn=1anconvergesif the sequence(sk)k2Nis convergent, and the seriesdivergesif the sequence(sk)k2Nis divergent.Since each partial sum is a finite sum of real numbers, eachskis very clearly defined; itis the sum of the firstkterms, the running tally as you sum the series one term at a time.The series is the limiting value, if such a limit exists.Remark 3.2.From the definition we note that the convergence or divergence of the seriesdoesn’t depend on which value ofnthe sum starts with; the seriesP1n=1an,1Xn=0an,1Xn=3an,and1Xn=2018anare either all convergent or all divergent. All that matters is the “tail” of theinfinite sequence (sn)n2Nof partial sums.Example: Geometric Series1Xn=0rn, wherer2Ris constant.This is the most important series, and it illustrates divergence and convergence very well.This is one of the only series where we have an explicit formula for the partial sums:sk=kXn=0rn=1-rk+11-r,k= 0,1,2,. . . .If you want to start the series withn= 1, then each term has a common factor ofr, so1Xn=1rn=r1Xn=0rn, andkXn=1rn=rk-1Xn=1rn=r1-rk1-r=r-rk+11-r.30
1Xn=1
If|r|<1, thenrk+1---!k!10, and so the partial sums converge, and the series isconvergent,1Xn=0rn= limk!1sk= limk!11-rk+11-r=11-r,and the limit value is explicitly known.Ifr >1,rn! 1(properly divergent), and sincesk> rkthe partial sums properlydiverge to infinity, the series is divergent. Ifr= 1, thenrk= 1 andsk=k+ 1 (the numberof terms) and the series is also properly divergent to infinity.Whenr <-1,rk+1is unbounded and diverges, so also doessk.(But not properly!|rk+1|---!k!10 but the values oscillate in sign.) A more interesting case isr=-1, for whichskalternates between 1 (keven) and 0 (kodd). So a series can diverge even if the partialsums remain bounded; there are divergent series which are not properly divergent to±1.For most convergent series we don’t know to what value the series converges to. This iswhere Cauchy sequences come in handy: a series converges if and only if its partial sumsform a Cauchy sequence!Theorem 3.3.The seriesP1n=1anconverges if and only if:8">0,9H2Nso thatpXn=m+1an<",8p > mH.(3.1)Proof.This is not difficult to prove, since the left-hand side of (3.1) can be rewritten interms of the partial sums,|sp-sm|=pXn=m+1an.So the condition (3.1) is exactly the statement that the partial sums (sk)k2Nform a Cauchysequence inR. By Theorem 2.32 they converge if and only if (3.1) holds.An immediate consequence comes from looking at a special case of (3.1), whenm+1 =p,and there is only one term in the sum,|ap|=pXn=pan<",8p > H.

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