Every vector in a Hilbert space has a Fourier representation wrt an orthonormal sequences?

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I'm reading Kreyszig's text, and there is a Theorem in section 3.5 stating:

Theorem: Let $(e_k)$ be an orthonormal sequence in a Hilbert space $H$. Then

1) If $\sum_{k=1}^\infty \alpha_k e_k$ converges then the coefficients $\alpha_k$ are the Fourier coefficients $\langle x, e_k \rangle$, where $x=\sum_{k=1}^\infty \alpha_k e_k$, so that $x = \sum_{k=1}^\infty \langle x, e_k \rangle e_k$.

2) For all $x \in H$ the series $\sum_{k=1}^\infty \langle x, e_k \rangle e_k$ converges (in the norm of $H$).

I would like to check my understanding here. I believe we can then say that, given some orthonormal sequence $(e_k)$ and a fixed $x \in H$, we have that $\sum_{k=1}^\infty \langle x, e_k \rangle e_k$ converges by part 2). Then, by part 1), we can write $x = \sum_{k=1}^\infty \langle x, e_k \rangle e_k$. So every $x \in H$ can be represented as $\sum_{k=1}^\infty \langle x, e_k \rangle e_k$? Is my logic correct here?

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Your conclusion is too strong. By (2) you can conclude that $\sum_{k=1}^{\infty}\langle x,e_k\rangle e_k$ converges to some $y$, but you cannot conclude that $y=x$. Then by (1) you know that $y=\sum_{k=1}^{\infty}\langle y,e_k\rangle e_k$. So you can conclude that $\sum_{k=1}^{\infty}\langle y,e_k\rangle e_k = \sum_{k=1}^{\infty}\langle x,e_k\rangle e_k$, which implies that $y-x$ is orthogonal to every $e_k$ for $1 \le k < \infty$.