How do you prove a sequence is a Cauchy sequence?

A sequence {an}is called a Cauchy sequence if for any given ϵ > 0, there exists N ∈ N such that n, m ≥ N =⇒ |an − am| < ϵ. |an − L| < ϵ 2 ∀ n ≥ N. Thus if n, m ≥ N, we have |an − am|≤|an − L| + |am − L| < ϵ 2 + ϵ 2 = ϵ.

How do you use Cauchy criterion to prove convergence?

If a sequence (xn) converges then it satisfies the Cauchy’s criterion: for ϵ > 0, there exists N such that |xn − xm| < ϵ for all n, m ≥ N. If a sequence converges then the elements of the sequence get close to the limit as n increases.

Is (- 1 N Cauchy sequence?

Think of it this way : The sequence (−1)n is really made up of two sequences {1,1,1,…} and {−1,−1,−1,…} which are both going in different directions. A Cauchy sequence is, for all intents and purposes, a sequence which “should” converge (It may not, but for sequences of real numbers, it will).

How do you prove Cauchy criterion?

The sequence xn converges to something if and only if this holds: for every ϵ > 0 there exists K such that |xn − xm| < ϵ whenever n, m>K. This is necessary and sufficient. for m, n ≥ K.

How do you prove a series converges?

If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the root test is inconclusive, and the series may converge or diverge. The ratio test and the root test are both based on comparison with a geometric series, and as such they work in similar situations.

How do you prove that every Cauchy sequence is convergent?

Let ϵ > 0. Choose N so that if n>N, then xn − a < ϵ/2. Then, by the triangle inequality, xn − xm = xn − a + a − xm < ϵ if m,n>N. Hence, {xn} is a Cauchy sequence.

Which of the following sequence is Cauchy sequence?

Cauchy sequences are intimately tied up with convergent sequences. For example, every convergent sequence is Cauchy, because if a n → x a_n\to x an​→x, then ∣ a m − a n ∣ ≤ ∣ a m − x ∣ + ∣ x − a n ∣ , |a_m-a_n|\leq |a_m-x|+|x-a_n|, ∣am​−an​∣≤∣am​−x∣+∣x−an​∣, both of which must go to zero.