Article directory
introduction
It’s time to tackle the tough series, but in fact, you don’t have to be so afraid of it. I didn’t learn it when I first came into contact with it a few years ago, but now I might be able to pass it.
1. Constant item series
1.1 Basic concepts
Constant term series—— Let { an } \{a_n\}{ an} is a constant column, which means∑ n = 1 ∞ an \sum_{n=1}^\infty a_n∑n=1∞anis the series of constant terms.
Convergence and divergence of constant term series —— S n = a 1 + a 2 + ⋯ + an S_n =a_1+a_2+\dots+a_nSn=a1+a2+⋯+anis the series ∑ n = 1 ∞ an \sum_{n=1}^\infty a_n∑n=1∞anThe partial sum of , if lim n → ∞ S n \lim_{n\to\infty}S_nlimn→∞SnExistence, called series ∑ n = 1 ∞ an \sum_{n=1}^\infty a_n∑n=1∞anConvergence, let the value of this limit be SSS , then the series is said to converge toSSS. _ If this limit does not exist, the series is said to diverge.
1.2 Basic properties
Characteristics 1 —— ∑ n = 1 ∞ un = A , ∑ n = 1 ∞ vn = B , 则 ∑ n = 1 ∞ ( un ± vn ) = ∑ n = 1 ∞ un ± ∑ n = 1 ∞ vn = A ± B . \sum_{n=1}^\infty u_n=A,\sum_{n=1}^\infty v_n=B,则 \sum_{n=1}^\infty (u_n\pm v_n)=\sum_ {n=1}^\infty u_n\pm\sum_{n=1}^\infty v_n=A\pm B.n=1∑∞un=A,n=1∑∞vn=B,butn=1∑∞(un±vn)=n=1∑∞un±n=1∑∞vn=A±B.
Adding a convergent series to a divergent series results in divergence.
But the addition of two divergent series does not necessarily diverge. Such as ∑ n = 1 ∞ ( − 1 ) n + ∑ n = 1 ∞ ( − 1 ) n − 1 = 0. \sum_{n=1}^\infty(-1)^n+\sum_{n=1} ^\infty(-1)^{n-1}=0.n=1∑∞(−1)n+n=1∑∞(−1)n−1=0.
Property 2 ——Assume ∑ n = 1 ∞ un = S , then ∑ n = 1 ∞ kun = k S . If k ≠ 0 , both converge and diverge. Suppose \sum_{n=1}^\infty u_n= S, then \sum_{n=1}^\infty ku_n=kS. If k\ne0, both have the same convergence and divergence.set upn=1∑∞un=S,butn=1∑∞k un=k S . If k=0,The convergence and divergence of the two are the same . Property 3——The increase, decrease, and change of the finite item in the series will not change the convergence and divergence of the series, but may change the sum of the series.
Property 4 —— If a series converges, any series after adding parentheses will also converge. Conversely, if the series with parentheses converges, the original series may not necessarily converge (that is, adding parentheses can increase the possibility of series convergence ).
For example, the series ∑ n = 1 ∞ ( − 1 ) n \sum_{n=1}^\infty (-1)^n∑n=1∞(−1)n diverges, but( − 1 + 1 ) + ( − 1 + 1 ) + … (-1+1)+(-1+1)+\dots(−1+1)+(−1+1)+... convergence.
Property 5 —— (Necessary condition for series convergence) If the series ∑ n = 1 ∞ an converges, then lim n → ∞ an = 0 ; otherwise, not necessarily. If the series \sum_{n=1}^\infty a_n converges, then \lim_{n\to\infty}a_n=0; otherwise not necessarily.If the seriesn=1∑∞anConvergence, thenn→∞liman=0;The opposite is not necessarily the case. Proof:LetS n = a 1 + a 2 + ⋯ + an S_n=a_1+a_2+\dots+a_nSn=a1+a2+⋯+an, from the convergence of the series, we can see that lim n → ∞ S n \lim_{n\to\infty}S_nlimn→∞Snexists, set its value to SSS。因为an = S n − S n − 1 a_n=S_n-S_{n-1}an=Sn−Sn−1 ,所以 lim n → ∞ a n = lim n → ∞ S n − lim n → ∞ S n − 1 = S − S = 0. \lim_{n\to\infty}a_n=\lim_{n\to\infty}S_n-\lim_{n\to\infty}S_{n-1}=S-S=0. n→∞liman=n→∞limSn−n→∞limSn−1=S−S=0. For series∑ n = 1 ∞ 1 / n \sum_{n=1}^\infty 1/n∑n=1∞1/ n , althoughlim n → ∞ 1 / n = 0 \lim_{n\to\infty}1/n=0limn→∞1/n=0 , but it diverges.
1.3 Two important series
1.3.1 p series
(1) Definition: in the form of ∑ n = 1 ∞ 1 / np \sum_{n=1}^\infty 1/n^p∑n=1∞1/nThe series of p is called ppp series, whenp = 1 p=1p=When 1 , say ∑ n = 1 ∞ 1 / n \sum_{n=1}^\infty 1/n∑n=1∞1/ n is the harmonic series.
(2) Convergence and divergence judgment
- When p ≤ 1 p \leq 1p≤1 o'clock,ppp -series diverges; in particular, harmonic series diverges;
- when p > 1 p > 1p>1 o'clock,ppThe p -series converges.
1.3.2 Geometric progression
(1) Definition: in the form of ∑ n = 1 ∞ aqn ( a ≠ 0 ) \sum_{n=1}^\infty aq^n(a\ne0)∑n=1∞aqn(a=0 ) is called a geometric series.
(2) Convergence and divergence judgment
- 当∣ q ∣ ≥ 1 |q| \geq∣q∣≥When 1 , the geometric progression diverges;
- 当 ∣ q ∣ < 1 |q| < 1 ∣q∣<When 1 , the geometric progression converges, and its value isS = aq 1 − q S=\frac{aq}{1-q}S=1−qaq
1.4 Positive term series and its convergence and divergence judgment
(1) The concept of positive term series
design ∑ n = 1 ∞ an \sum_{n=1}^\infty a_n∑n=1∞anis a series of constant terms, if for all nnn有n ≥ 0 a_n \geq 0an≥0 , which is called a positive term series.
The biggest feature of the positive series is the partial sum sequence { S n } \{S_n\}{ Sn} monotonically increasing, there are two cases:
- { S n } \{S_n\} { Sn} There is no upper bound, thenlim n → ∞ S n = + ∞ \lim_{n\to\infty}S_n=+\inftylimn→∞Sn=+ ∞ , at this time the positive term series diverges;
- There exists M > 0 M>0M>0 , makingS n ≤ M S_n \leq MSn≤M , thenlim n → ∞ S n \lim_{n\to\infty}S_nlimn→∞Snexists, the positive term series converges.
(2) Convergence check method of positive term series
I'll type it into a form in Word and post a screenshot.
The general idea of judging the convergence and divergence of positive series is as follows:
(1) Judging whether the series meets the necessary conditions for convergence, if not, the series diverges;
(2) If the general term of the series is the difference between two adjacent terms of the series, the definition method is generally used;
(3) For non-specific positive term series whose general terms meet certain conditions, generally use the nature and discriminant method of the convergence and divergence of the series;
(4) For the general and specific series of positive terms, the specific convergence method is generally used: if the general term contains factorial, the ratio verification method is generally used; if the general term contains nnFor n powers, the root value convergence method is generally used; if the general item contains logarithms, the integral convergence method is generally used; for other cases, the ratio convergence method is generally used.
1.5 Interleaved series and its convergence method
(1) The concept of alternating series
∑ n = 1 ∞ ( − 1 ) nun \sum_ { n=1}^\infty (-1)^nu_n∑n=1∞(−1)n unor∑ n = 1 ∞ ( − 1 ) ( n − 1 ) un \sum_{n=1}^\infty (-1)^{(n-1)}u_n∑n=1∞(−1)(n−1)unis an alternating series, where un > 0. u_n>0.un>0.
(2) Leibniz's Appraisal Method
设∑ n = 1 ∞ ( − 1 ) ( n − 1 ) un \sum_{n=1}^\infty (-1)^{(n-1)}u_n∑n=1∞(−1)(n−1)unIt is an interleaved series, if the series satisfies the following two conditions:
(1) { u n } \{u_n\} { un} Monotonically decreasing; (2)lim n → ∞ un = 0 , \lim_{n\to\infty}u_n=0,limn→∞un=0,
Nominal number ∑ n = 1 ∞ ( − 1 ) ( n − 1 ) un \sum_{n=1}^\infty (-1)^{(n-1)}u_n∑n=1∞(−1)(n−1)unConverge, and its sum does not exceed u 1 . u_1.u1.
The two conditions for the convergence of an alternating series are sufficient but not necessary.
1.6 Absolute convergence and conditional convergence of series
(1) Basic concepts
若∑ n = 1 ∞ ∣ un ∣ \sum_{n=1}^\infty |u_n|∑n=1∞∣un∣Convergence , then it is said that∑ n = 1 ∞ un \sum_{n=1}^\infty u_n∑n=1∞unAbsolutely restrained.
若∑ n = 1 ∞ a \sum_{n=1}^\infty u_n∑n=1∞un收敛,However∑ n = 1 ∞ ∣ un ∣ \sum_{n=1}^\infty |u_n|∑n=1∞∣un∣发 dispersion, also known as∑ n = 1 ∞ un \sum_{n=1}^\infty u_n∑n=1∞unConditional convergence.
(2) The relationship between absolute convergence and conditional convergence
若∑ n = 1 ∞ a \sum_{n=1}^\infty u_n∑n=1∞unAbsolute convergence, then ∑ n = 1 ∞ un \sum_{n=1}^\infty u_n∑n=1∞unconvergence, and vice versa.
Some notes and summary:
(1) Use the following formulas to help memorize the judgment of convergence: adding parentheses increases the possibility of series convergence; the faster the general item tends to 0, the greater the possibility of series convergence; adding absolute values increases the possibility of convergence .
(2) The first order of judging the convergence of the constant term series:
Step 1: Check whether the necessary conditions for series convergence are met;
The second step: see if the convergence and divergence can be judged according to the definition;
The third step: determine the specific series type, and use the respective convergence method to judge according to the above.
(3) The necessary and sufficient condition for the convergence of positive term series is ∑ n = 1 ∞ u 2 n − 1 \sum_{n=1}^\infty u_{2n-1}∑n=1∞u2n − 1 _Sum ∑ n = 1 ∞ u 2 n \sum_{n=1}^\infty u_{2n}∑n=1∞u2 nAll converge.
(4)若∑ n = 1 ∞ a \sum_{n=1}^\infty u_n∑n=1∞un收敛,则∑ n = 1 ∞ ( un + un + 1 ) \sum_{n=1}^\infty (u_n+u_{n+1)}∑n=1∞(un+un+1)Must converge (adding parentheses increases the likelihood of convergence).
(5)若∑ n = 1 ∞ a \sum_{n=1}^\infty u_n∑n=1∞un收敛、∑ n = 1 ∞ un 2 \sum_{n=1}^\infty u_n^2∑n=1∞un2Not necessarily convergent; but if ∑ n = 1 ∞ un \sum_{n=1}^\infty u_n∑n=1∞unConvergent positive term series, ∑ n = 1 ∞ un 2 \sum_{n=1}^\infty u_n^2∑n=1∞un2Must converge.
write at the end
The theoretical part of the constant term series is here, and we will continue to learn the content of the power series later.