Section 1.3 Convergence Tests We learned in the previous section that for any geometric series, we know exactly whether it converges and the value it converges to. For most series, however, there is no exact formula similar to that for geometric series. In applications, one often approximates a series by one of its partial sum if the series is convergent. In other words, N 1 an sN = an for an appropriate (large) number N. n=0 ¼ n=0 X X Two questions need to be answered before using this approximation: (1) is the series convergent? (2) how to select the number N? If the series is divergent, this approximation doesn’t make any sense. For a convergent series, we could select small N if the series converges very "fast". Otherwise, for a slowly convergent series, the number N must be a very large number. In this section we shall discuss these questions. We shall study six tests that can be used determine whether a series is convergent. Test 1: Integral Test Consider non-negative series 1 an, an 0. n=0 ¸ X Let us recall that an improper integral is de…ned as n 1 f (x) dx = lim f (x) dx, n Z1 !1 Z1 and is called convergent if the limit exists and is …nite. We now assume that f (x) is non-negative and decreasing function for x > 0 and that an = f (n) (i.e., an 0 and is a decreasing sequence). ¸ 1 From the …gure above, …rst rectangle area = a 1 = a curved area under y = f (x) over [0, 1] 1 ¢ 1 · 2nd rectangle area = a2 curved area under y = f (x) over [1, 2] · 3rd rectangle area = a curved area under y = f (x) over [2, 3] 3 · nth rectangle area = an curved area under y = f (x) over [n 1, n] · ¡ Sum of …rst n rectangular area = a1 + a2 + ... + an curved area over [0, n] · Therefore, n an f (x) dx, · n 1 Z ¡ and sn = a1 + a2 + ... + an 1 2 n n f (x) dx + f (x) dx + ... + f (x) dx = f (x) dx. · 0 1 n 1 0 Z Z Z ¡ Z On the other hand, if we look at the same graph of y = f (x) again with di¤erent dashlined rectangles, 2 then we see that, in a similar manner, …rst rectangle area = a1 curved area under y = f (x) over [1, 2] ¸ 2nd rectangle area = a curved area under y = f (x) over [2, 3] 2 ¸ 3rd rectangle area = a3 curved area under y = f (x) over [3, 4] ¸ n+1 n+1 an f (x) dx, sn f (x) dx. ¸ ¸ Zn Z1 We summarize the above analysis by following inequalities n+1 n f (x) dx an f (x) dx (1) n · · n 1 ¡ Z n+1 Z n f (x) dx s f (x) dx. · n · Z1 Z0 Note that the above inequalities can be easily proved using abstract al- gebraic analysis (v.s. lengthy graphic display) as follows. Since an = f (n) , and the function f (x) is decreasing, we have f (x) an if n x · · a f (x) if x n n · · 3 Integrating on the …rst inequalities over (n, n + 1) and the second one over (n 1, n) , respectively, we arrive at ¡ n+1 n+1 n+1 f (x) dx andx = f (x) dx an n · n ) n · Z n Z n Z n andx f (x) dx = an f (x) dx. n 1 · n 1 ) · n 1 Z ¡ Z ¡ Z ¡ These are precisely formula (1). 1 Using (1), we see that if 0 f (x) dx is convergent, then n R 1 sn f (x) dx f (x) dx < · · 1 Z0 Z0 i.e.., sn is bounded. Note that since an 0, ¸ sn = sn 1 + an sn 1 ¡ ¸ ¡ is always increasing. Thus, sn is bounded increasing sequence which implies 1 s is convergent = a is convergent. n ) n n=1 X 1 On the other hand, if an is convergent, then n=1 P 1 s a < is bounded, n · n 1 n=1 X and by (1) n+1 n+1 1 f (x) dx sn is bounded = f (x) dx = lim f (x) dx is convergent. · ) n Z1 Z1 !1 Z1 We conclude the above discussion as follows: 4 Integral Test: Suppose that an is non-negative and decreasing as described above. Then 1 1 an is convergent i¤ (if and only if) f (x) dx is convergent. n=1 1 X Z Also from i+1 i f (x) dx ai f (x) dx, i · · i 1 Z Z ¡ we conclude that i+1 i 1 1 1 1 1 f (x) dx = f (x) dx ai f (x) dx = f (x) dx n+1 i · · i 1 n Z i=n+1 Z i=n+1 i=n+1 Z ¡ Z X X X (2) This last inequality is often referred to as 1 Reminder Estimate: Suppose an is convergent, and de…ne the n=1 reminder term Rn of the series as P 1 1 Rn = ai = an+1 + an+2 + ... = an sn. i=n+1 Ãn=1 ! ¡ X X Then the inequalities (2) read as 1 1 f (x) dx Rn f (x) dx. (3) · · Zn+1 Zn These inequalities are used to make prediction on possible errors if we use a partial sum sn as an approximation to a series. Example 3.1. (p series) The series ¡ 1 1 , p > 0. np n=1 X is called p-series. Discuss convergence or divergence of p-series. Solution: Consider f (x) = 1/xp. This function is decreasing and positive for x > 0. Thus the integral test can be applied to conclude that 1 1 is convergent IFF (if and only if) np n=1 X 1 1 dx is …nite. xp Z1 5 For p = 1, 6 1 p 1 1 p dx = x¡ dx = x ¡ xp 1 p Z Z ¡ n 1 p 1 1 1 1 n ¡ 1 if 1 p < 0 dx = lim dx = lim = ¡1 p ¡ . xp n xp n 1 p 1 p 8 1 !1 1 !1 ¡ ¡ if 1 p > 0 Z Z µ ¡ ¡ ¶ < 1 ¡ For p = 1, : 1 1 dx = dx = ln x xp x Z Z n 1 1 1 dx = lim dx = lim (ln (n) ln 1) = . x n x n ¡ 1 Z1 !1 Z1 !1 We conclude that 1 1 dx is convergent IFF 1 p < 0 (or p > 1). xp ¡ Z1 Thus, 1 1 p series is convergent IFF p > 1. np ¡ n=1 X ln n Example 3.2. Determine whether 1 is convergent. n=1 n Solution: Consider function P ln x y = . x Since its derivative 1 ln x y0 = ¡ 0 if x > 3 > e, x2 · this function is decreasing and positive for x > 3. So the integral test may be applied. Since by substitution u = ln x 1 ln x du = dx u 1 (ln x)2 dx =x xdu = udu = u2 = , x x 2 2 Z Z Z 6 we …nd n 2 2 1 ln x ln x (ln x) (ln 3) dx = lim dx = lim = . x n x n 2 ¡ 2 1 Z3 !1 Z3 !1 Ã ! By the integral test, 1 ln n = is divergent. n n=3 1 X Test #2: Comparison Test. Consider two non-negative series and suppose that 0 an bn, n = 0, 1, 2, ... · · Then, the same relation holds for their partial sums: relations a0 b0 · a1 b1 · a b 2 · 2 .... imply n n 0 s = a b = S . · n i · i n i=0 i=0 X X Furthermore, since an, bn 0, both partial sum sequences sn and Sn are increasing. ¸ f g f g 1 If bi is convergent, then i=0 P 1 s S b (…nite number). n · n · i i=0 X So, in this case the partial sum sequences sn is increasing and bounded above, and therefore it is convergent. We jufst pgroved 1 1 (A) If bi is convergent, then so is ai. i=0 i=0 P P 7 1 Similarly, if ai is divergent, then sn is unbounded from above, i.e., i=0 P 1 sn . Consequently, Sn and bi is divergent, too. ! 1 ! 1 i=0 1 P1 (B) If ai is divergent, then so is bi. i=0 i=0 The coPmparison test may be writtenPand memorized conveniently as 1 1 0 an bn = an bn, · · ) n=0 · n=0 X X in which the last inequality is interpreted as 1 1 1 1 If bn < ( bn is …nite), then an bn < . n=0 1 n=0 n=0 · n=0 1 X X X X 1 1 1 1 If an = , then = an bn = bn = . n=0 1 1 n=0 · n=0 ) n=0 1 X X X X To apply Comparison Test to a non-negative series, we often …rst make an educated guess whether a given series is convergent or divergent. If we want 1 to verify this series is convergent, then we call it an and …nd a larger and n=0 known to be convergent series (likely either a geoPmetric series or p-series) as 1 bn. Vice versa, if we guess the series is divergent and want to prove it, n=0 P 1 1 then we call it bn and try to …nd a smaller divergent series as an. n=0 n=0 Example 3P.3. Discuss convergence for P 1 1 (a) n ; n=0 2 + 1 P1 5 (b) 2 ; n=0 2n + 4n + 1 ln n (c) P1 ; n=1 n + 1 P1 5n (d) 2 .