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Unformatted text preview: MATH 118, LECTURES 22 & 23: SERIES: RATIO & ROOT TESTS 1 Series 1.1 Limit Ratio Test Consider the convergence of the series summationdisplay n =1 1 n ! . This can be handled by the comparison test, taking an appropriate choice of bounding function. We will, however, use an alternative intuition. We can relate successive terms c n +1 and c n very easily: we have c n +1 = 1 n c n parenleftbigg or c n +1 c n = 1 n parenrightbigg . There was another series we could relate in this fashion, which was the geometric series where the difference between the successive terms given by the relationship c n +1 = rc n . We recall that geometric series converged if  r  < 1 and diverged if  r  1. But how do we relate our factor 1 /n to the geometric series multiplicative factor r ? For geometric series, r was fixed for all values of n , so how do we know which value of n to pick to get our r ? We recall that the convergence or divergence of series is really dependent only on the large values of n , so that we can can take the limit as n . Since 1 /n goes to 0 as n , we suspect this series in convergent. In fact, this is justified by the following result: Proposition 1.1 (Limit Ratio Test) . Suppose that c n > and lim n c n +1 c n = L. Then: 1 1. n =1 c n converges if L < 1 . 2. n =1 c n diverges if L > 1 (or if lim n c n +1 /c n = ). 3. n =1 c n may converge or diverge if L = 1 . In this case, we have lim n c n +1 /c n = lim n 1 /n = 0 so the series converges. Justification: We can handle this by cases. In the first two cases, we will use the formal definition of a limit to construct a geometric sequence which bounds the original series (from above in the first case, and below in the second)we will then use the comparison test. From the formal definition of a limit, we have that lim n c n +1 /c n = L implies that, for all epsilon1 > 0 there exists an N 1 such that vextendsingle vextendsingle vextendsingle vextendsingle c n +1 c n L vextendsingle vextendsingle vextendsingle vextendsingle < epsilon1 = ( L epsilon1 ) c n < c n +1 < ( L + epsilon1 ) c n , for n N....
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 Spring '09
 Soares

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