Geometric Series and the Ratio and Root Test

Geometric Series and the Ratio and Root Test James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University September 5, 2017 Outline Geometric Series The Ratio Test The Root Test Examples A Taste of Power Series

Theorem Geometric Series: n=0 r n converges if 1 < r < 1 to 1/(1 r). n=0 r n diverges if r ( 1, 1). (1): We show n=0 r n converges if 1 < r < 1 to 1/(1 r). We already know the partial sum Sn = 1 + r +... + r n = (1 r n+1 )/(1 r) form a previous induction argument. We also know that r n+1 0 as n since 1 < r < 1. Thus, n=0 r n converges to 1/(1 r). (2): If r = 1, we have the series 1 + 1 + + 1 +... which has partial sums Sn = n. The partial sums are unbounded and so the series does not converge. If r = 1, we have seen already that n=0 ( 1)n diverges by oscillation. If r > 1, we have Sn = 1 + r +... + r n = (r n+1 1/(r 1) and we see limn Sn = since r n+1 and r 1 > 0. So this series diverges. If r < 1, the limn r n does not exist and so by the n th term test, the series must diverge. Comment: note if 1 < r < 1 (1) n=1 r n = 1/(1 r) 1 (2) n=2 r n = 1/(1 r) 1 r (3) n=3 r n = 1/(1 r) 1 r r 2 So (1) n=1 (1/3)n = 1/(1 (1/3)) 1 = 3/2 1 = 1/2 (2) n=2 (1/3)n = 1/(1 (1/3)) 1 (1/3) = 1/2 1/3 = 1/6 (3) n=3 r n = 1/(1 (1/3)) 1 (1/3) (1/9) = 1/6 1/9 = 1/(18)

Theorem The Ratio Test: Let n=1 an be a series with positive terms. and assume limn an+1/an = ρ. (1) if 0 ρ < 1, the series converges. (2) if ρ > 1, the series diverges. (3) if ρ = 1, we don t know. (1): We assume 0 ρ < 1. Then, we see ρ < (1 + ρ)/2 < 1. Let the value (1 + ρ)/2 be the r in the geometric series n=1 r n which converges since 0 < r < 1. We will use the comparision test to show our series converges. Let ɛ = (1 ρ)/2 which is a nice positive number. Then, ɛ + ρ = (1 + ρ)/2 = r < 1 Since limn an+1/an = ρ, for this ɛ, there is an N so that n > N = an+1 ρ < ɛ = 1 ρ an 2 Thus, an+1/an < ɛ + ρ = (1 + ρ)/2 = r < 1 for all n > N. We see an+2 < r an+1 an+3 < r an+2 < r (r an+1) = r 2 an+1 an+4 < r an+3 < r (r 2 an+1) = r 3 an+1 for l > 1.. an+l < r l 1 an+1 = an+1 l+n r r N+1

Now let k = N + l and c = an+1/r N+1. We have shown therefore that ak < cr k for k > N + 1. It is easy to see the series c k=n+2 r n converges by the limit comparison test, so by the comparison test, we have shown k=n+2 ak converges which also tells us k=1 ak converges. (2): If ρ > 1, choose ɛ = (ρ 1)/2 > 0 as ρ > 1. Then, there is an N so that n > N = an+1 ρ < ɛ = ρ 1 an 2 This implies for n > N that 1 + ρ = ρ ɛ < an+1 < ρ + ɛ 2 an Thus, letting r = (1 + ρ)/2 > 1, we have an+2 > r an+1 an+3 > r an+2 > r (r an+1) = r 2 an+1 an+4 > r an+3 > r (r 2 an+1) = r 3 an+1. an+l > r l 1 an+1 = an+1 l+n r r N+1 for l > 1. Let C = an+1/r N+1 and k = N + l. Then, we have ak > C r k for all k > N + 1. Now apply the comparison test. Since k=n+2 r k diverges since r > 1 and we know C k=n+2 r k diverges, by the comparison test k=n+2 ak diverges. (3): If ρ = 1, the series could converge or diverge.

(i) the series n=1 1/n2 converges but ρ = 1. (ii) the series n=1 1/n diverges but ρ = 1. So if ρ = 1, we just don t know. We can say more: Theorem The Ratio Test for Absolute Convergence: Let an+1 n=1 an be a series with nonzero terms. Assume limn an converges to the value ρ. (1) if 0 ρ < 1, the series converges. (2) if ρ > 1, the series diverges. (3) if ρ = 1, we don t know. (1): If 0 ρ < 1, we can apply the previous theorem to see n=1 an converges. But if a series converges absolutely, it also converges. (2): If ρ > 1, the arguments in the proof above show us there is an N so we can write ak Cr k for k > N + 1 for r = (1 + ρ)/2 > 1. Now if k=n+2 ak converged, this implies ak 0. But if that were true, ak 0 too. But the inequality ak Cr k with r > 1 says the sequence ( ak ) can not converge. Thus, k=n+2 ak must also diverge. Of course, this tells us n=1 an also diverges. (3): The same examples as before show us why we can t conclude anything about convergence or divergence of the series if ρ = 1.

Theorem The Root Test: Let n=1 an be a series with non-negative terms. Assume limn (an) 1/n exists and equals ρ. (1) if 0 ρ < 1, the series converges. (2) if ρ > 1, the series diverges. (3) if ρ = 1, we don t know. (1): We assume 0 ρ < 1. Choose ɛ = (1 ρ)/2 > 0. Then there is an N so that n > N = (an) 1/n ρ < ɛ = (1 ρ)/2 We can rewrite this: for n > N (1 ρ)/2 < (an) 1/n ρ < (1 ρ)/2 = (an) 1/n < (1 + ρ)/2 < 1 Let r = (1 + ρ)/2 < 1. (an+1) 1/(N+1) < r = an+1 < r N+1 (an+2) 1/(N+2) < r = an+2 < r N+2 (an+l) 1/(N+l) < r = an+l < r N+l. for l 1. Letting k = N + l, we have ak < r k for k > N. Since 0 < r < 1, the series k=n+1 r k converges and so by comparison, so does k=n+1 ak. This shows k=1 ak converges. (2): Now ρ > 1. This argument is very similar to the one use for the proof of the ratio test.

Let ɛ = (ρ 1)/2 > 0. Then there is an N so that We can rewrite this as n > N = (an) 1/n ρ < ɛ = (ρ 1)/2 (1 ρ)/2 < (an) 1/n ρ < (ρ 1)/2 = (an) 1/n > (1 + ρ)/2 > 1 for n > N. Let r = (1 + ρ)/2. Then, we have the inequality an > r n for all n > N. Since n=n+1 r n diverges since r > 1, by the comparison test, we know n=n+1 an diverges and so the original series diverges too. (3): If ρ = 1, we have limn n 1/n = 1. Look at the series n=1 1/n. We know this diverges and ρ = 1. Now look at the series n=1 1/n2. This converges and limn (n 2 ) 1/n = limn n 2/n. But limn n 2/n = limn (n 1/n ) 2. Since the function f (x) = x 2 is continuous, we also know limn (n 1/n ) 2 = ( limn n 1/n ) 2 = 1. So we just don t know whether the series converges or diverges if ρ = 1. We can restate this result for series whose terms are not necessarily non-negative.

Theorem The Root Test for Absolute Convergence: Let n=1 an be a series. Assume limn an 1/n exists and equals ρ. (1) if 0 ρ < 1, the series converges. (2) if ρ > 1, the series diverges. (3) if ρ = 1, we don t know. (1): If 0 ρ < 1, the previous theorem tells us n=1 an converges absolutely implying convergence. (2): If ρ > 1, the earlier arguments imply is an N so ak r k for k > N for r = (1 + ρ)/2 > 1. If ak converged, then ak 0 and ak 0 k=n+1 too. But ak r k with r > 1 says the sequence ( ak ) can not converge. Thus, k=n+1 ak must also diverge and so the original series diverges too. (3): The same examples as before show us why we can t conclude anything about convergence or divergence of the series if ρ = 1. Let s summarize what we know: (1) We can check convergence with comparison tests. If the terms are not non-negative, we can test for absolute convergence which will imply convergence. (2) The ratio and root test are also good tools. We can check for absolute convergence if the terms are not non-negative to see if the series converge. But the root test requires that none of the terms are zero.

Example What is the sum of (3/4) 5 + (3/4) 6 +... + (3/4) n +...? This is a geometric series with r = 3/4 missing the first 5 terms. So the sum of the series is 1/(1 (3/4)) 1 (3/4) (3/4) 2 (3/4) 3 (3/4) 4. Note the five terms we subtract could be written 1 + (3/4) + (3/4) 2 + (3/4) 3 + (3/4) 4 = (1 (3/4) 5 )/(1 (3/4)) but this is not necessarily an improvement! Example (1): What is the sum of k=1 (1/(3k) 1/(4k))? To do this we have to write the inner part as a single fraction. We find ( 1/(3k) 1/(4k) ) = k/(12k 2 ) = 1/(12k) k=1 which diverges as it is multiple of the harmonic series. (2): Does k=5 ( (2k3 + 3)/(k 5 4k 4 ) ) converge? This looks like the series 1/k 2 so a good choice would be the limit comparison test. But that gives k=1 lim k (2k3 + 3)/(k 5 4k 4 ) /(1/k 2 = lim k (2k5 + 3k 2 )/(k 5 4k 4 ) = 2 k=1 Thus, since the limit is positive and finite, our series converges.

Example Note the ratio test will fail. We find (2(k + 1) 3 + 3)/((k + 1) 5 4(k + 1) 4 ) lim k (2k 3 + 3)/(k 5 4k 4 ) = lim k (2(k + 1) 3 + 3) (k 5 4k 4 ) ((k + 1) 5 4(k + 1) 4 ) (2k 3 + 3) 2k 8 + lower power terms of k = lim k 2k 8 + lower power terms of k = 1 and so the ratio test fails. Of course, it would be insane to apply the root test here! Example Does k=1 1/( (2k + 1)2k ) converge? We ll try the ratio test here: 1/( (2(k + 1) + 1)2 k+1 ) (2k + 1)2 k 2k + 1 lim k 1/( (2k + 1)2 k = lim = lim ) k (2k + 3)2k+1 k (2k + 3)2 The limit is 1/2 and hence, this series converges by the ratio test.

A series of the form n=0 anx n is called a power series centered at x = 0. Note no matter what the numbers an are, this series always converges at x = 0. Let s assume the terms an are not zero and that limn an+1/an = ρ > 0. We can check for the convergence and divergence of this series using the ratio test. lim an+1x n+1 n anx n = lim an+1 n x = ρ x an By the ratio test, the series converges when ρ x < 1 and diverges when ρ x > 1. So we have (1) The series converges when 1/ρ < x < 1/ρ (2) The series diverges when x > 1/ρ or x < 1/ρ (3) We don t know what happens at x = ±1/ρ. The number 1/ρ is called R, the radius of convergence of the power series. Hence, this power series defines a function f (x) locally at x = 0 with radius of convergence R > 0. Any function f (x) which has a power series which matches it locally at x = 0 is said to be analytic at x = 0. The Derived series obtained from f (x) = n=0 anx n is the series n=0 n anx n 1. We are still assuming limn an+1/an = ρ > 0, so we can check the convergence of this series using the ratio test too. lim (n + 1) an+1x n ( ) n n anx n 1 = lim an+1 n + 1 n lim x = ρ x an n n Hence, this series converges when x < R = 1/ρ also. We see the derived series defines a function g(x) on the circle ( R, R). An obvious question is this: Is f (x) = g(x) on ( R, R)? That is does ( ) f (x) = anx n = n=0 ( ) anx n = n anx n 1 = g(x)? n=0 Note this is an interchange of limit order idea. Is it possible to interchange the limit called differentiation with the limit called series? When we can do this, this is called differentiating the series term by term. n=0

We will find that we can differentiate and integrate a power series term by term as many times as we like and the radius of convergence remains the same! These are the kinds of series we construct using Taylor polynomials and the resulting series is called a Taylor Series expansion of f (x) at x = 0. We have a long way to go to be able to prove this sort of theorem. But this is the kind of manipulation we must do to use power series to solve ODEs with non constant coefficients! Another kind of series that is very useful is a0 + n=1 ( an cos(nπx/l) + bn sin(nπx/l) ) which is called a Fourier series. Note its building blocks are sin and cos functions rather than powers of x. Homework 6 6.1 Determine if n=1 n!/2n converges. 6.2 Determine if n=1 ( ( 1)n (2n + 5) )/n! converges. 6.3 Determine if n=1 2/(n3 + 5n 2) converges. 6.4 Determine for what values of x the series n=0 ( 1)n x 2n /( (2n)! ) converges and for what values of x it diverges.