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

Question

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!

Answer 1

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.