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

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 $$(C[a, b], ||•||_1)$$?

• if the normed space is not complete, it may have one absolutely convergent series which is not convergent. Nov 25, 2021 at 7:02
• 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 Nov 25, 2021 at 8:32
• @Kavi Rama Murthy : $\sum (x^n-x^{n+1})$ is convergent in $(C[0, 1], ||•||_1)$ Nov 26, 2021 at 21:21

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 $$n_1 such that $$\|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.