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I know that in a Banach space $(X, ||•||) $ absolutely convergent series is convergent.

But if a normed space $(X,||•||)$ is not complete, then there exists atleast one absolutely convergent series which is not convergent.

I am able to show that $(C[a, b], ||•||_1) $ is not complete.

Where, $C[a, b]=\{f: [a, b] \to \mathbb{R} : f \text{ is continuous }\}$

And, $||f||_1 = \int_{a}{^b}{|f(t)| dt}$

My Question: how to construct an absolutely convergent series which is not convergent in the space $(C[a, b], ||•||_1) $?

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    $\begingroup$ if the normed space is not complete, it may have one absolutely convergent series which is not convergent. $\endgroup$ Nov 25, 2021 at 7:02
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    $\begingroup$ Thank you for down voting. I am new in this subject. So I don't have enough experience playing with these kind of problem. The question may be non sense but it's my doubt. So, I have to clear it. And that's why i posted this question. Criticism is always expected for my mistakes but leave a comment so that I can't repeat my mistake next time. Thanks $\endgroup$ Nov 25, 2021 at 8:32
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    $\begingroup$ @Kavi Rama Murthy : $\sum (x^n-x^{n+1})$ is convergent in $(C[0, 1], ||•||_1) $ $\endgroup$
    – perroquet
    Nov 26, 2021 at 21:21

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Here is a general argument.

Since you are able to show that a normed space $X$ is not complete, I assume you reached to a point where you found a Cauchy sequence $(x_n)$ that does not converge, correct?

Since $(x_n)$ is Cauchy, find $n_1\in\mathbb{N}$ such that if $n,m\ge n_1$ then $\|x_n-x_m\|<\frac{1}{2}$. Then, find $n_2>n_1$ such that if $n,m\ge n_2$ then $\|x_n-x_m\|<\frac{1}{2^2}$. Continuing this process, we obtain indices $n_1<n_2<\dots$ such that $\|x_{n_{k+1}}-x_{n_k}\|<\frac{1}{2^k}$ for all $k\in\mathbb{N}$.

Set $y_{k}=x_{n_{k+1}}-x_{n_k}$ for all $k\ge1$. Then, $$\sum_k\|y_k\|\le\sum_k\frac{1}{2^k}<\infty$$ On the other hand, the partial sums are $s_K=\sum_{k=1}^Ky_k=x_{n_{K+1}}-x_{n_1}$, so, if the series $\sum_ky_k$ converges, then the sequence $\{x_{n_k}-x_{n_1}\}_{k=1}^\infty$ converges and thus the sequence $\{x_{n_k}\}$ converges. But a Cauchy sequence having a convergent subsequence is also convergent to the limit of the subsequence. This cannot be, since we know that $(x_n)$ does not converge.

One can come up with this by looking at the standard proof of the fact that completeness is equivalent to the implication of series convergence by absolute convergence.

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    $\begingroup$ @s.g. this argument is an example provider, that's my point. You say in your post that you proved the space is not complete, so you have a specific non convergent cauchy sequence right? Just apply the argument I'm showing here to construct an example of a non convergent series. $\endgroup$ Nov 27, 2021 at 9:05

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