# How to prove the 'uniform summability' of a Cauchy sequence?

I have an exercise given by the teacher and I'm pretty sure that this proof is not hard, but I don't have idea how to approach it. I have to prove the 'uniform summability' (this name was used by professor) of Cauchy sequence in $l^2$:
for a Cauchy sequence $(x^{(n)})$ in $l^2$ and $\epsilon >0$, prove there exists $K >0$ such that for all $n$ $\sum_{j=K}^{\infty} |x_{j}^{(n)}|^2 < \epsilon$

Do you have some hints or ideas, how to start this proof?

Thanks in advance for any help!

The best place to start is always by writing down what we know.

First:

$(x^{(n)})$ is Cauchy, so given $\epsilon > 0$, $\exists N > 0$ such that $\forall m,n\ge N$, $$||x^{(m)} - x^{(n)}||<\epsilon$$

Second:

$(x_n)$ is in $\ell_2$ - so $$||x^{(n)}||= \left(\sum_{j=1}^\infty|x^{(n)}_j|^2\right)^\frac12$$converges and hence, for each $n$, $\exists K_2(n)$ such that $$\sum_{j=K_2(n)}^\infty|x^{(n)}_j|^2<\epsilon$$

Let $K = \max\{N, K_2(1), K_2(2), \ldots, K_2(N)\}$

Hint: Use the above and the fact that given $n > N$, $$||x^{(n)}||\le||x^{(N)}||+||x^{(n)}-x^{(N)}||$$to prove the result.

• Thanks for the answer! I'm not sure if the 'hint' will be helpful - you said that this inequality is true for the n's greater than the N defined in the Cauchy condition; but I have to prove that it works for all n. – yarpen Aug 3 '14 at 21:09
• Have a look at the way I've defined $K$ - the hint was to show you how $K$ works for every $n$, however it will most certainly work for $n \le N$ – Mathmo123 Aug 3 '14 at 21:11
• I'm very sorry for stupid questions, it's my first experience with this branch of mathematics and I don't see many things. I guess that this definition of K gives me correctness for n $\leq$ N and I'm supposed to prove that it will hold for any n greater than N? And it is a property of each convergent series that we can choose a K such that sum from K to $\infty$ will be arbitrary small, right? – yarpen Aug 3 '14 at 21:30
• That's correct. The point is, that for any given $N$ we can choose $K$ such that the property holds for all $n\le N$. But to get the result for all $n$, we need to use the Cauchy property. – Mathmo123 Aug 3 '14 at 21:36