Proving Convergence of a Sequence Using Cauchy Criterion

[andyfeynman]

Homework Statement

Show that the sequence {x_n}:
x_n := (2^1/1 - 1)² + (2^1/2 - 1)² + ... + (2^1/n - 1)² is convergent.

Homework Equations

The Attempt at a Solution

If n > m,
|x_n - x_m| = (2^1/n - 1)² + (2^1/(n-1) - 1)² + ... + (2^1/(m+1) - 1)²
< (2^1/n)² + (2^1/(n-1))² + ... + (2^1/(m+1))²
< (2^1/(m+1))² + (2^1/(m+2))² + ...
= 4^1/m
Let ɛ > 0. We choose N such that 4^1/N < ɛ for all n > m > N.
Then |x_n - x_m| < 4^1/N < ɛ for all n > m > N.

 

[mfb]

Let ɛ > 0. We choose N such that 4^1/N < ɛ for all n > m > N.

Which N would you choose for ɛ=1?

 

[andyfeynman]

Which N would you choose for ɛ=1?

Just forgot it. I made a very stupid mistake.
But I came up with the idea of letting b_k = 2^1/k - 1.
This means b_k² < 4[k(k-1)] for all k > 2.
Therefore,
x_n = 1 + b₂² + ... + b_n²
< 1 + 4/[1(2-1)] + ... 4/[n(n-1)]
= 1 + 4(1 - 1/2) + ... + 4[1/(n-1) - 1/n]
= 5 - 4/n
< 5
Since x_n is monotone increasing and bounded, it is convergent.
But is there any way to do it using the Cauchy criterion?

 

[mfb]

This means b_k² < 4[k(k-1)] for all k > 2.

That step is certainly not trivial.

I would use something like 2^1/k < 1 + c/k for some c.

Cauchy criterion: Probably, but I don't see how it would help.