A relatively short inductive proof of Steinitz Theorem can be founded in this paper. Here we present a sketch of the proof, which is based on 3 lemmas whose proof is left to the reader. First we recall the formulation of the result.

**Theorem.** A series $\sum_{n=1}^\infty x_n$ in a finite-dimensional Banach space $X$ is "good" if and only if for any linear functional $f:X\to\mathbb R$ the series $\sum_{n=1}^\infty f(x_n)$ is "good".

*Proof.* This theorem will be proved by induction on the dimension of $X$. The theorem is trivially true for Banach spaces of dimension $\le 1$.
Assume that for some number $d$ the theorem has been proved for all Banach spaces of dimension $<d$.

Fix a Banach space $X$ of dimension $d$. We can assume that $X=\mathbb R^d$ and the norm $\|\cdot\|$ of $X$ is generated by the standard inner product $\langle\cdot,\cdot\rangle$.

Let $\sum_{n=1}^\infty x_n$ be a series in $X$ such that for every linear functional $f:X\to\mathbb R$ the series $\sum_{n=1}^\infty f(x_n)$ is "good". We lose no generality assuming that $x_n\ne 0$ for all $n$.

**Lemma 1.** $\lim_{n\to\infty}x_n=0$.

**Definition.** A point $s$ on the unit sphere $S=\{x\in X:\|x\|=1\}$ is called a *divergence direction* of the series $\sum_{n=1}^\infty x_n$ if for any neighborhood $U\subset S$ of $x$ the set $\mathbb N_U=\{n\in\mathbb N:\frac{x_n}{\|x_n\|}\in U\}$ is infinite and $\sum_{n\in\mathbb N_U}\|x_n\|=\infty$.

Let $D$ be the (closed) set of all divergence directions of $(x_n)_{n\in\mathbb N}$.

If $D=\emptyset$, then by the compactness of the sphere $S$, $\sum_{n=0}^\infty \|x_n\|<\infty$ and the series $\sum_{n=1}^\infty x_n$ is convergent, so is "good".

So, we assume that $D$ is not empty.
Using Hahn-Banach Theorem it can be shown that the convex hull $conv(D)$ of $D$ contains zero.

Let $D_0$ be a subset of smallest cardinality in $D$ such that $0\in conv(D_0)$.

It can be shown that the set $D_0$ is affinely independent and zero is contained in the interior of the simplex $conv(D_0)$.

Let $X_0$ be the linear hull of the set $D_0$ and $X_1$ be the orhogonal complement of $X_0$ in the Euclidean space $X$.
So, $X=X_0\oplus X_1$.

For $i\in\{0,1\}$ let $pr_i:X\to X_i$ be the orthogonal projections of $X$ onto $X_i$.

**Lemma 2.** There exists a subset $\Omega\subset \mathbb N$ such that

1) $\sum_{n\in\Omega}\|pr_1(x_n)\|<+\infty$;

2) for any $s\in D$ and a neighborhood $U\subset S$ of $s$ the set $\Omega_U=\{n\in\Omega:\frac{x_n}{\|x_n\|}\in U\}$ is infinite and $\sum\limits_{n\in\Omega_U}\|pr_0(x_n)\|=\infty$.

**Lemma 3.** There exists a constant $C$ (dependent on $D_0$) such that for any $x\in X_0$, $\varepsilon>0$, and a finite set $\Omega_0\subset \Omega$ there exists a finite set $F\subset \Omega\setminus \Omega_0$ such that

1) $\|x-\sum_{n\in F}pr_0(x_n)\|<\varepsilon$;

2) $\sum_{n\in F}\|x_n\|\le C\max\{\|x\|,\varepsilon\}$.

Since $X_1$ has dimension $<d$, we can apply the inductive assumption and conclude that the series $\sum_{n\in\mathbb N\setminus\Omega}pr_1(x_n)$ is "good", which means that for some permutation $\sigma_1$ of $\mathbb N\setminus\Omega$ the series $\sum_{n\in\mathbb N\setminus\Omega}pr_1(x_{\sigma_1(n)})$ is convergent in the Banach space $X_1$.

Then using Lemma 3, we can extend the permutation $\sigma_1$ of $\mathbb N\setminus\Omega$ to a permutation $\sigma$ of $\mathbb N$ such that $\sum_{n=1}^\infty pr_0(x_{\sigma(n)})=0$.

Since $\sum_{n\in\Omega}\|pr_1(x_n)\|<+\infty$, the series
$\sum_{n\in\Omega}pr_1(x_{\sigma(n)})$ converges and then the series $\sum_{n\in\mathbb N}pr_1(x_{\sigma(n)})=\sum_{n\in\mathbb N\setminus\Omega}pr_1(x_{\sigma(n)})+\sum_{n\in\Omega}pr_1(x_{\sigma(n)})$ converges, too.

Then the series $\sum_{n\in\mathbb N}x_{\sigma(n)}=\sum_{n\in\mathbb N}pr_0(x_{\sigma(n)})+pr_1(x_{\sigma(n)})$ also converges and hence the series $\sum_{n\in\mathbb N}x_n$ is "good".