# The ratio and root test also work but the ratios and

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The ratio and root test also work, But the ratios and roots have to be computed correctly! . Now fl fl fl x n +1 1+ x 2 n +2 fl fl fl fl fl fl x n 1+ x 2 n fl fl fl = | x | (1 + x 2 n ) 1 + x 2 n +2 . It is quite obvious that if | x | < 1, then lim sup n →∞ | x | (1 + x 2 n ) 1 + x 2 n +2 = lim n →∞ | x | (1 + x 2 n ) 1 + x 2 n +2 = 0 . Since 0 < 1, the series converges if | x | < 1. The limsup is not so easy to calculate if | x | > 1 and I would expect a Calculus 1 student to have some difficulty with it. But lim sup n →∞ | x | (1 + x 2 n ) 1 + x 2 n +2 = lim n →∞ | x | 2 n +1 ( 1 + 1 x 2 n ) x 2 n +2 ( 1 + 1 x 2 n +2 ) = lim n →∞ 1 | x | fl fl 1 + 1 x 2 n fl fl ( 1 + 1 x 2 n +2 ) = 1 | x | < 1 . And we have convergence for | x | > 1. The test fails for | x | = 1, but divergence is obvious (as done before). The root test also is easy to apply, if one knows what to do with n 1 + x 2 n . One probably can’t get away from once again dividing into the cases | x | > 1, | x | < 1 and | x | = 1. In this last case, the test is inconclusive, so one has to proceed as before. But it is not incredibly hard to prove that lim n →∞ n s fl fl fl fl x n 1 + x 2 n fl fl fl fl = lim n →∞ | x | n 1 + x 2 n = 0 if | x | < 1, 1 if | x | = 1, 1 / | x | if | x | > 1. 4. Assume f n : R R for n = 1 , 2 , . . . are such that { f n } is equicontinuous. (a) (20 points) If { f n } converges pointwise to some function f ; i.e., if f ( x ) = lim n →∞ f n ( x ) exists for all x R , prove f is continuous. Solution. Arzela-Ascoli does 90% of the work, except that Arzela Ascoli is valid only on compact domains. But continuity is a local property, so no real problem. There is also a quickie proof that bypasses Arzela-Ascoli. Proof. Let x 0 R , we have to prove f is continuous at x 0 . Select any closed and bounded interval such that x 0 is in its interior; for example, let a = x 0 - 1 , b = x 0 + 1. The restrictions of the functions f n to the interval [ a, b ] still form an equicontinuous sequence. If we want to be super perfectly rigorous we can define g n = f n | [ a,b ] ; then obviously { g n } is equicontinuous, converges point wise (necessarily) to f | [ a,b ] and because it converges pointwise and is equicontinuous it is also uniformly bounded on [ a, b ] (See below for a proof of this fact, which has already been proved before in this course). Thus, By Arzela-Ascoli, { g n } has a uniformly convergent subsequence on [ a, b ], the limit can only be (once more) f . Thus f | [ a,b ] , as the uniform limit of a sequence of continuous functions (members of an equicontinuous set are trivially continuous) must be continuous on [ a, b ]. Because x 0 is in the interior of [

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