The ratio and root test also work But the ratios and roots have to be computed

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 [