Infinite Series Summary

Transcription

1 Infinite Series Summary () Special series to remember: Geometric series ar n Here a is the first term and r is the common ratio. When r <, the series is convergent, and we have: ar n = a r When r, the series is divergent. The harmonic series n is a famous divergent series with a n = /n 0. This example reminds us that a n 0 alone cannot guarantee the convergence of the corresponding series a n. The p series n p When p >, the series is convergent. When p, the series is divergent. Note that when p = the p series has another name: the harmonic series. The alternating series ( ) n b n with b n > 0 When b n+ b n and b n 0, it is convergent. When b n \ 0, it is divergent. So the alternating harmonic series: is convergent. ( ) n n

2 2 Note: () When the ratio of a geometric series is negative, it is also an alternating series. In this case, it is much more productive to treat it as geometric series. (2) For alternating series, be aware of the difference between a n and b n. It is a very helpful habit to write down the b n explicitly. (3) To verify b n is decreasing: Suppose that when we substitute x for n in b n we get a function f(x), where b n = f(n). You can show f(x) is a decreasing function by determining that its derivative f (x) is on an interval (a, ). (Usually the endpoint a will be or 2.) (2) Convegence Tests for a series nth term test: a n : If lim n a n does not exist or exists but the limit is not zero, then the series diverges. Note that this test cannot be used to show convergence. Integral Test: Only for a n 0 Suppose that replacing n by x in a n results in a function f(x), where f(n) = a n. Form the improper integral I = I converges = a n converges I diverges =. f(x)dx. Simple Comparison Test: Only for a n 0 You must use an appropriate positive series b n for which you know its convergence. (a) a n b n and (b) a n b n and b n converges = b n diverges = a n converges Limit Comparison Test: Only for a n 0 Choose an appropriate positive series b n for which you know its convergence. a n lim = L > 0 = series n b n a n and b n have the same convergency. a n

3 Ratio Test: Useful when factorials appear in a n. a n+ Compute lim = L n a n L < = series a n absolutely converges L > = series Root Test: Useful when a n contains factors raised to the power n. n Compute lim an = L n L < = series a n absolutely converges L > = series (3) The Race to Infinity. How to guess the convergence of a given series in the form: a n where a n = c n and lim d c n = lim d n = with c n and d n both positive? n n n First we note that the convergence of a n somewhat depends on the relative speed of how fast c n and d n approach infinity. For example, n 3 + approaches faster than n 2 n and n do, but n 3 + converges and diverges. The reason? The relative speeds of approaching n 2 n 3 + are different: n 3 + is faster relative to x as to x 2. The following is an incomplete list of f(n) with the speed approaching infinity from the slowest to the fastest: lnn n p r n n! n n where p > 0 and r > are constants. f(n) (The expression f(n) g(n) means lim n g(n) = 0.) Try to guess the convergency of series: lnn n, p nln p n, n p r n nlnn(ln(lnn)) p, r n, n!, and Test your guess for each! What if c n faster than d n? The original series diverges. n! n n 3

4 4 (4) Absolute and conditional convergence: Series a n is absolutely convergent series Series series a n is conditionally convergent a n is convergent, but series a n is convergent. a n is divergent. From above definitions we know that the only series that could be conditionally convergent is the series having infinitely many negative terms and infinitely many positive terms. In this book, that is the alternating series. Thus, the only series that could be the candidate for conditionally convergent is the alternating series (in this text book). Any series that is tested to be convergent by using the tests listed in 2 must be absolutely convergent. (Why?) Examples: () (2) lnn n p p : Use the integral test: Integration by parts (u = lnx, dv = x p dx) gives t lnxx p ( t dx = (lnx)x p+ t 2 x p+ ) p+ x dx 2 t ( = (lnx)x p+ t ) 2 p+ 2 x pdx. When t, if p >,x p+ t 2 = t p 2 and the integral p 2p 2 x pdx converges too; if p <, they both diverge. So it is convergent when p > and divergent if p <. p = : Use the Integral test: (see Example 5, p.62 in the textbook) or else compare it with the harmonic series. It will be divergent. Conclusion: The original series convergent when p > and divergent when p. n! n n Use the ratio test: a n+ (n+)!n n n n = lim = lim a n n n!(n+) n+ n (n+) = lim n n ( n+ The last equation comes from a famous limit: lim n n lim n 2 t ( n+ ) n = n e ) n = e, which can be

5 shown by H lopital s rule. Since /e <, according to the test the original series converges. (3) Determine whether the series converges or diverges. If it 8 converges find its exact sum. We recognize this as an example of both an alternating series and a geometric series. However, since the problem also asks us the exact sum, we should treat it as geometric series. 5