Question

A sequence (xn) converges quadratically to x if there is some Q ∈ R such that |xn − x| ≤ Q/n^2

for all n ∈ N. Prove directly that if (xn) converges quadratically, then it is also Cauchy.

Answer 1

◆ Cauchy sequence: A sequence is called a Cauchy sequence if the terms of the sequence eventually all become arbitrarily close to one another.

i.e. given € > 0 ( epsilon) , there exists a natural number N , such that, if m, n > N , then |am - an|<€ .

● Now to prove { xn} is Cauchy , let {xn} converges quadratically to some x.

Then there is some Q in R , such that ,

Now , let €/2 > 0 ( € means epsilon), then there exists a natural number N , such that , Q/n2 < € , for all n > N.

Therefore the above expression becomes,

Or,.   

Now let , m, n be two natural numbers grater than N. Then,

We have to show , | xm - xn| < € ( epsilon)

So,

  

Since m , n are arbitrary , therefore for any positive epsilon , we get a natural number N , such that,

Hence the sequence {xn} is Cauchy.