Complete orthonormal sequence $\iff$ finite linear combinations are dense in a Hilbert space How would I prove, that the following statements are equivalent?

*

*$\{v_n\}_{n=1}^{\infty}$ is a complete orthonormal sequence.

*$\lim\limits_{N \to \infty} || v - \sum_{n = 1}^{N} \langle v, v_n\rangle  v_n|| = 0$ for every $v \in V$, where $V$ is a Hilbert space.

This is my attempt:
1 $\implies$ 2: If $\{v_n \}$ is a complete orthonormal sequence, then for every $v \in V$:
$$v = \sum\limits_{n=1} ^{\infty} \langle v, v_n\rangle  v_n =  \lim_{N \to \infty} \sum_{n = 1}^{N} \langle v, v_n\rangle  v_n||. $$
Is this correct? I am struggling with the other implication. It seems quite obvious but I do not quite know if this is correct:
2 $\implies$ 1: If $\lim\limits_{N \to \infty} || v - \sum_{n = 1}^{N} \langle v, v_n\rangle  v_n|| = 0$ for every $v \in V$, it means that
$$ v = \sum_{n=0}^{\infty} \langle v, v_n\rangle  v_n $$ for every $v \in V$ which is exactly the definition of CONS.
 A: Completeness of an orthonormal system $S$ means that it cannot be extended, i.e. there is no nonzero element orthogonal to all the elements of $S.$ In order to show $2\implies 1,$ assume there is $v$ such that $v\perp v_n$ for every $n.$ Then
$$\|v\|=\left \|v-\sum_{k=1}^n\langle v,v_n\rangle v_k\right \|\underset{n}{\longrightarrow} 0$$ Hence $v=0,$ which means $\{v_n\}_{n=1}^\infty $ is complete.
A: If by "complete orthonormal sequence", you mean orthonormal basis (or something equivalent, such as a maximal subset of orthonormal vectors), then (1) implies (2) is a standard result in functional analysis (Your argument is short of details. Basically, you need to show $\sum_{i=1}^\infty (v, v_i)v_i$ converges and $v-\sum_{i=1}^\infty (v, v_i)v_i$ is orthogonal all of $v_i$'s hence must be $0$), while (2) doesn't imply (1). In particular, they are not even necessarily orthogonal.
This is the well-known phenomenon that gave birth to the notion of "frame" instead of "base". To give a concrete example, in $\mathbb R^2$ (or $\mathbb C^2$, whether the Hilbert space is complex is no big deal for the following to work), consider the three vectors that distribute uniformly on the unit circle: $$w_1=(1, 0), w_2=(-\frac{1}{2}, \frac{\sqrt 3}{2}), w_3=(-\frac{1}{2}, -\frac{\sqrt 3}{2})$$ Now it's easy to verify $$\sum_{i=1}^3(v, w_i)w_i=\frac{3}{2}v$$ Hence the family of vectors $v_i=\sqrt{\frac{2}{3}}w_i$ satisfies (2), but not (1), as $v_i$'s are clearly not orthogonal.
To get an example in the infinite dimensions, we may group a true orthonormal basis into couples, and just find the corresponding $v_1, v_2, v_3$ for each couple.
I suspect what you're asked to show is given an orthonormal sequence $\{v_i\}$, then (1) It's complete and (2) $\sum_{i=1}^\infty (v, v_i)=v$ are equivalent.
