Question

Exercise 2.4.5: Suppose that a Cauchy sequence {xn} is such that for every M ∈ N, there exists a k ≥ M and an n ≥ M such that xk < 0 and xn > 0. Using simply the definition of a Cauchy sequence and of a convergent sequence, show that the sequence converges to 0.

Answer #1

Let (xn) be Cauchy in (M, d) and a ∈ M. Show that the
sequence
d(xn, a) converges in R. (Note: It is not given that
xn converges to a.
Hint: Use Reverse triangle inequality.)

Use the definition of a Cauchy sequence to prove that the
sequence defined by xn = (3/2)^n is a Cauchy sequence in R.

Given that xn is a sequence of real numbers. If (xn) is a
convergent sequence prove that (xn) is bounded. That is, show that
there exists C > 0 such that |xn| less than or equal to C for
all n in N.

Suppose that every Cauchy sequence of X has a convergent
subsequence in X. Show that X is complete.

Let xn be a sequence such that for every m ∈ N, m ≥ 2 the
sequence limn→∞ xmn = L. Prove or provide a counterexample: limn→∞
xn = L.

If (xn) ∞ to n=1 is a convergent sequence with limn→∞ xn = 0
prove that
lim n→∞ (x1 + x2 + · · · + xn)/ n = 0 .

Let 0 < θ < 1 and let (xn) be a sequence where
|xn+1 − xn| ≤ θn for n
= 1, 2, . . ..
a) Show that for any 1 ≤ n < m one has |xm −
xn| ≤ (θn/ 1-θ )*(1 − θ m−n ).
Conclude that (xn) is Cauchy
b)If lim xn = x* , prove the following error in
approximation (the "error in approximation" is the same as error
estimation in Taylor Theorem) in t:...

Let {Xn} be a sequence of random variables that follow a
geometric distribution with parameter λ/n, where n > λ > 0.
Show that as n → ∞, Xn/n converges in distribution to an
exponential distribution with rate λ.

Suppose {xn} is a sequence of real numbers that converges to
+infinity, and suppose that {bn} is a sequence of real numbers that
converges. Prove that {xn+bn} converges to +infinity.

Exercise 1. Suppose (a_n) is a sequence and f : N --> N is a
bijection. Let (b_n) be the sequence where b_n = a_f(n) for all n
contained in N. Prove that if a_n converges to L, then b_n also
converges to L.

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