Suppose Σ XK is conditionally convergent, and define k=1 ak = max {xk, 0} and bk = min {xk, 0} ∞ Notice that xk = a + bk and |xk| = ak - bk. Show that ✗ ak and ☐ bk are both divergent. k-1 K-1
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- . Let gn = nχ[1/n,2/n] and g = 0. Show thatZ 20g 6= limn→∞ Z 20gn.Does the sequence (gn) converge uniformly to g? Does the Monotone Convergence Theoremapply? Does Fatou’s Lemma apply?(c) Carefully construct a nested family of subsequences (fm,k), and show howthis can be used to produce a single subsequence of (fn) that convergesat every point of A.a) Suppose (an) is Cauchy and that for every k ∈ N, the interval (−1/k, 1/k) contains at least one term of (an). Can we say that (an) converges to 0? Either show that it does or give a counter-example.
- Let a > 0 and X1 = √a. Define the sequence Xn = √(a + Xn-1), n ≥ 1.Show that (Xn)n is convergent and determine its limit.Suppose that ak ≥ 0 and that ak^1/3→ a as k → ∞.Prove that sigma,k from 1 to ∞,akx^k converges absolutely for all |x| < 1/a if a ≠ 0 and for all x ∈ R if a = 0Prove that the sequence {cn} converges to c if and only if the sequence {cn- c} converges to 0.
- Apply the Ratio Test to determine convergence or divergence, or state that the Ratio Test is inconclusive.∑n=1∞enn!ρ=limn→∞|an+1an|= (Enter 'inf' for ∞.) ∑n=1∞enn! is:A. convergentB. divergentC. The Ratio Test is inconclusiveSuppose that F(u) denotes the DFT of the sequence of f(x)={1, 2, 3, 4}? What is the value of F(14)? (Hint: DFT periodicity)Find suitable coefficients (an) so that the resulting power seriessigma anxn has the given properties, or explain why such a request is impossible. (c) Converges absolutely for all x ∈ [−1, 1] and diverges off of this set.
- Use the Cauchy Condensation test to prove that ∑ n = 2 to ∞ 1/( n (ln(n))^ p)) converges if p > 1 and diverges if p ≤ 1. (Make sure you verify that the hypothesis of the Cauchy Condensation test are met)The fourth term of the Taylor series of f(x)=lnx at x=2 isProve that if {sn} converges to L and L > 0, then there exists a number N such that sn > 0 for n > N.