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B) Prove that n=1 an is convergent. If this is the case, we have solved our problem. If not, we have got a problem. Something may be rescued by giving a better estimate in a), which means that one in one’s first try has given a too crude estimate. 1 The series n=1 because � � � �1 1 � � � n2 sin nx� < n , and even �∞ 1 �∞ 1 1 sin nx has the two majoring series n=1 and n=1 2 , n n n2 � � � �1 1 � � � n2 sin nx� ≤ n2 . com 53 Calculus 3b General series; tricks and methods in solutions of problems The first majoring series is divergent (hence no conclusion, because the estimate is too crude), and the second one is convergent, hence the trigonometric series is uniformly convergent.
Are (slowly) convergent for α > 1 and (slowly) divergent for α ≤ 1. We here say that the convergence/divergence is slow, in order to express that we have absolutely no chance in calculating the value of the series by using pocket calculators or MAPLE. For instance, if we let a pocket calculator add the first 106 terms of the (divergent) harmonic series above, it is easy by applying the integral test that the sum has barely passed 21, so one would be tempted wrongly to conclude that the series is convergent.
G. MAPLE. It is remarkable that an old proof which a long time ago sunk into oblivion now again can be used with the new advanced computer programs at hand. ♦ The proofs of the first and third case above rely on the important fix point theorem. This theorem will be treated more thoroughly in the next chapter. g. f1 (x) = 0 or f2 (x) = x, where we assume that f1 and f2 are continuous. In the former equation we shall find the real zeros of f (x), and in the latter equation we shall find fix points of the function f , if zeros or fix points do exist.