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Convergent and Divergent Infinite Series: Mathematical Induction and the Binomial Theorem

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Convergent and Divergent Infinite Series: Mathematical Induction and the Binomial Theorem CONVERGENT AND DIVERGENT INFINITE SERIES: MATHEMATICAL INDUCTION AND THE BINOMIAL THEOREM In the last unit we examined techniques for establishing the limits of sequences and used a few examples of infinite series to illustrate the increased complexity of problems that deal with infinite processes. In this unit we will examine infinite series in more detail and develop techniques for establishing their sums. Infinite Series: The Ratio Test Infinite Series: The Polynomial Quotient Test Infinite Series: Combination Tests Principle of Math Induction The Binomial Theorem Infinite Series: The Ratio Test As stated in the last unit, infinite series may actually sum to a final value. Their sums can be infinite, or they can be inconclusive. When a series sums to a final value, then the series “converges” or is “convergent”. When a series sums to infinity or is inconclusive, then the series “diverges” or is “divergent”. Ratio Test for Infinite Series: Let an and an+1 be two consecutive terms of a positive series. a Suppose lim n+1 = r where r ∈\ . Then the series converges if r <1; diverges if r >1and the n→∞ an series may or may not converge if r =1. It is important to note that the value of ""r is not the sum of the series. This value only tells you that the series converges but not to what value. Example #1: Use the ratio test to determine if the following series converges or diverges. 12 3 4 n ++ + +.......... + + ...... 3 9 27 81 3n Step #1: Find the general expression for the an and an+1 terms of the series. (Use techniques from previous units to assist in this process when the general pattern is not n provided.) In this example an = : Therefore the an+1 term is, 3n n +1 an+1 = . 3n+1 Step #2: Find the ratio of the two terms and simplify. n +1 a n+1 nn++13nn 1 3 n + 1 n+1 ==⋅=⋅=3 nn+1 annnn n 333⋅ 3 3n Step #3: Find the limit of the ratio. n +11 lim ==r n→∞ n⋅33 (Clearly the denominator and numerator increase at the same rate as n →∞ because the “+1” term contributes nothing to the result at ∞ ) 1 Step #4: Conclusion: Since r = <1 the series converges. 3 Infinite Series: The Polynomial Quotient Test Polynomial Quotient Test for Convergence (PQT) Suppose Ao and Bo are polynomial expressions of the forms: kk−−12 k 2 0 PAoo( )=++++++ ax o ax12 ax ....... a k−− 2 x a k 1 xax k nn−−12 n 2 0 Qoo( B )=++++++ bx o bx12 bx ....... b n−− 2 x b n 1 x bx n where nk, ∈\ : ab, ∈\ . PQ Then, for ooand the series converges for: x nnx Paaaa a ok12++++3 4 . + nknAaxkn−−1234 − ax kn lim xxxxxx==k 00 + = k→∞ Qbb00bb b onBn bx0012++++3 4 . + bx xxxxxxnn1234 provided kn−≤0; otherwise the series diverges. Although this theorem appears quite complicated, its application to series convergence is quite useful and not too difficult to determine as the following examples illustrate; but first, a special notation is described to indicate series. To denote a series of any form, we use the Geek letter “Sigma” which appears as: k ∑ aaaank=+++123..... a n=1 ∞ If the series is infinite we have ∑ aaaank=+++123..... a + ...... n=1 The bottom notation "1"n = , indicates where the series begins and the top notation, (k or ∞), where the series ends. For example: 8 11111 =+++ which when evaluated = 0.1038 . ∑ 22222 n=5 n 5678 ∞ ∑ (1)2222222...−n ⋅=−+−+−+ which we learned in a previous unit diverges. n=0 Now that we have notation for series, we can use the Polynomial Quotient Test (PQT) in the following examples. Determine if the following series converge or diverge. ∞ 32nn2 + Example #1: ∑ 2 n=1 45n + 2 2 Step #1: By the PQT PAo ()= 3 n+ 2 n and QBo ()= 4 n+ 5 and km==2 . (For the highest exponent on ""n in each polynomial – use ""m for the exponent on QBo ( ) so that there is no confusion). PA() 32nn2 2 a) Find o = +=+3 . nnn222 n QA() 45n 2 5 b) Find o =+=+4 . nnn222 n 2 PA() 2 o 3+ 2 Step #2: Find lim nn= . QB() 5 n→∞ o 4 + n 2 n 2 2 5 Clearly as n →∞ “ + ” and “ + ” each come closer to zero. Therefore, at ∞ the n n 2 30+ 3 expression becomes = ; hence, the series converges. 40+ 4 ∞ 267nn2 +− Example #2: ∑ 3 n=1 5422nn++ 2 3 Step #1: PA0 ()=+− 2 n 6 n 7 QB0 ()= 5 n++ 4 n 22 ⇒= km2 and =3 Step #2: Since km< , we will divide both numerator and denominator by n3 which is the highest power term in the denominator which is stated in the theorem as: PA()o x n even though the highest power on PAo ( ) is k . QBo () x n The result of this division is: 26 7 +− ∞ 23 nn n ∑ 422 n=1 5 ++ nn23 Clearly as n →∞ every term in the expression except the “5” in the denominator approaches zero which gives ∞ 000++ 0 ∑ ==0 . Therefore the series converges. n=1 500++ 5 ∞ 46n 2 − Example #3: Find the value of . ∑ 42 n=1 nn+ 51− 2 Step #1: PAo ()=− 4 n 6 ⇒=k 2 42 QBo ()=+ n 5 n − 1 ⇒=m 4 However ,the presence of the "" symbol calls for a bit more analysis. Recall that 11 given the inequality ab>⇒ <. ab 11 Consider then nn42+−>51 n 4 ⇒ < nn42+−51 n 4 111 Which therefore implies that <=. 2 nn42+−51 n 4 n Hence m = 2 and not m = 4 as stated above, or, stated differently: PA() 46n 2 − 6 6 o 44−− m 222 Step #2: Find nn=== n n QBo () nn42+−51 nn 42 +− 51 51 m 1+− n 4 24 n 4 n nn 6 4 − 2 40− 4 Now find: limn = == 4 n→∞ 51 100+− 1 1+− nn24 Therefore the series converges. ∞ 95− n 4 Example #4: Find the convergence or divergence of: . ∑ 32 n=1 65nnn+ + 4 Step #1: PAo ()=− 9 5 n ⇒= k 4 32 QBo ()=++ 6 n 5 n n ⇒= m 3 Since k ≤m, the series diverges. or PA() 9 o − 5n 33 nn= , and QB() 5 1 o 6 ++ n3 nn2 9 − 5n 3 05−−nn 5 lim n ===−∞ 51 n→∞ 6 ++ 600++ 6 nn2 Therefore the series diverges. Infinite Series: Combination Tests The Ratio Test can be used in combination with the Polynomial Quotient Test (PQT) in the following manner. ∞ 27n + Example #1: Find the convergence or divergence of; . ∑ n n=1 2(n + 1) th 27n + Step #1: For the Ratio Test, the n term is an = and the n +1 term is 2(n n + 1) 2(nn++ 1) 7 2 ++ 2 7 2 n + 9 ===an+1 2(nnn+++111nnn++ 11)2( + 2)2( + 2) 29n + a 2(n+1 n + 2) 2(n nn++ 1)(2 9) ⇒ Comparing : n+1 == 27n + n+1 annn 2(++ 2)(27) 2(n n + 1) 2n (nn22++ 11 9) nn ++ 11 9 = = 2n ⋅++ 2(2nn22 11 14) 4 n ++ 22 n 28 2 2 Step #2: PAo ()=+ n 119 n + and QBo ()=++ 4 n 2228 n ⇒=km =2 Step #3: PA() 11 9 o 1++ 22 nnn= QB( ) 22 28 o 4 ++ n 2 nn2 11 9 1++ 2 100++ 1 ⇒==lim nn 22 28 n→∞ 4 ++ 400++ 4 nn2 Therefore the series converges. Example #2: Determine if the following series converges. n ⎛⎞2 2 ⎜⎟(5n ) 5 a = ⎝⎠ n 65n + By the Polynomial Quotient Test km= 2, =1, therefore the series would diverge. n ⎛⎞2 However the presence of the ⎜⎟term may have an effect on this outcome, so again the ⎝⎠5 Ratio Test comes into play. n n+1 ⎛⎞2 2 ⎛⎞2 2 ⎜⎟(5n ) ⎜⎟ (5(n + 1) ) 5 5 Step #1: a = ⎝⎠ a = ⎝⎠ n 65n + n+1 6(n ++ 1) 5 Step #2: Simplify an+1 . n n ⎛⎞⎛⎞22 2 ⎛⎞⎛⎞22 2 ⎜⎟⎜⎟(5(nn+ 2+ 1)) ⎜⎟⎜⎟(5nn+ 10+ 5) 55 55 a = ⎝⎠⎝⎠ = ⎝⎠⎝⎠ n+1 611n + 611n + a Step #3: Find the Ratio of n+1 . an nnn 22⎛⎞22 ⎛⎞ 2 22⎛⎞ 2 ⎜⎟(5nn++ 10 5) ⎜⎟ (5 n ) ⎜⎟(5nn++ 10 5) 55 5 55 65n + ⎝⎠÷ ⎝⎠ = ⎝⎠ × n 611nn++ 65 611n + ⎛⎞2 2 ⎜⎟(5n ) ⎝⎠5 2 (5nn2 ++ 10 5)(6 n + 5) (2nn2 + 4++ 2)(6 n 5) 5 = 5nn232 (6++ 11) 30 n 55 n Multiplying the numerator using algebra techniques gives the expression: (2nn232++ 4 2)(6 n += 5) 12 n + 34 n + 32 n + 10 a 12nnn32+++ 34 32 10 n+1 = 32 annn 30+ 55 Step #4: Use the PQT km= = 2 . 34 32 10 12 +++ 2312 2 lim nn n= = 55 n→∞ 30 + 30 5 n Therefore the series converges. Up to this point, all that has been introduced is to test whether an infinite series converges or diverges; but, no actual sum has been calculated. As has been noted over the last three units, many series do sum to a finite value that can be calculated. However techniques, which find actual sums, can be quite complicated and are better left for another course. For the remainder of this unit, we will examine two topics that play an important role in higher mathematics and extend the techniques and concepts learned thus far. These topics are the Principle of Math Induction and the Binomial Theorem. Principle of Math Induction For the past three units, we have often found single expressions that calculate the “nth” term in a sequence or expressions that sum a series. Yet when dealing with infinite processes, it is sometimes difficult to be sure if the expression chosen actually represents all terms even as n becomes quite large or infinite.
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