1.10. INDEPENDENT PROJECTS 27 Exercise 1.10.16. (i) If N N and z = 1, show that SN = N n=0 zn = 1−zN+1 1−z . (ii) If |z| 1, show that limn→∞ zn = 0. (iii) If |z| 1, show that limn→∞ zn does not exist. Theorem 1.10.17. Consider the geometric series defined by a complex number z. If |z| 1, then the series converges. If |z| 1, then the series diverges. Proof. This follows from the exercise above. Exercise 1.10.18. (i) What can you say if |z| = 1? (ii) Suppose that a series n=1 an converges. Show that limn→∞ an = 0. The property limn→∞ an = 0 does not ensure that the series n=1 an converges. The most useful example is given above where an = 1/n. In this case, S1 = 1, S4 2, and it is easy to check that S2n n for n N. It follows that the series n=1 1/n diverges. The series S = n=1 1/n is often called the harmonic series. We have just proved that this series diverges. Exercise 1.10.19. (i) Let SN = N n=1 1/n. Show that, for N 2, SN is never an integer. (ii) Show that, by suitably eliminating an infinite number of terms, the remaining subseries can be made to converge to any positive real num- ber. Exercise 1.10.20. (i) If s R and s 1, show that 1/ns converges. (ii) If s R and s 1, show that ∑n=1 1/ns diverges. (iii) For which s R does the series ∑=1n p prime 1/ps converge? Definition 1.10.21. A series n=1 an of complex numbers converges absolutely if the series n=1 |an| converges. Proposition 1.10.22. If n=1 an converges absolutely, then n=1 an converges Proof. This follows from the fact that | n k=m+1 ak| n k=m+1 |ak|. The converse to Proposition 1.10.22 is false and is shown by the example n=1 (−1)n+1/n. This series converges since | n k=m+1 (−1)k+1/k| 1/m. However, as we have seen above, the series does not converge absolutely. There are various tests to determine if a series converges. These include the comparison test, the ratio test, and the root test. The comparison test is often very useful, but its use depends on knowing ahead of time a series which converges. Theorem 1.10.23 (Comparison test). Suppose an 0 for n N and suppose n=1 an converges. If bn C satisfies |bn| an for all n, then the series n=1 bn converges absolutely and hence converges.
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