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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!

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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.

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  • $\begingroup$ 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. $\endgroup$ – yarpen Aug 3 '14 at 21:09
  • $\begingroup$ 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 $n \le N$ $\endgroup$ – Mathmo123 Aug 3 '14 at 21:11
  • $\begingroup$ 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? $\endgroup$ – yarpen Aug 3 '14 at 21:30
  • $\begingroup$ 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. $\endgroup$ – Mathmo123 Aug 3 '14 at 21:36

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