Are all rearrangement sums of a conditionally convergent vector series a subspace?

Q: Is the set of rearrangement sums of a conditionally convergent vector series always a subspace?

No. In general it is not closed under addition or scalar multiplication, so it need not be a linear subspace of R^d.

Q: What changes if the series is absolutely convergent?

Then every rearrangement has the same sum. The set of rearrangement sums becomes a singleton, which is a linear subspace only in the special case where that sum is the zero vector.

Question

1. Consider a conditionally convergent series of vectors $\sum_{n=1}^{\infty} v_n$ in $\mathbb{R}^d$ equipped with the standard Euclidean norm. Determine, with proof, whether the set of sums of all possible convergent rearrangements of this series is necessarily a linear subspace of $\mathbb{R}^d$.

Acemy AI · Accepted Answer

> **Key concept**: This is one of those questions where the scalar case tempts you into a quick analogy, but vector series behave a bit more rigidly. The structure of the set of rearrangement sums matters more than the series itself.

### Step by step
Let

$$S=\left\{ \sum_{n=1}^\infty v_{\sigma(n)} : \sigma 	ext{ is a permutation and the rearranged series converges} ight\}.$$

We ask whether $S$ is necessarily a linear subspace of $\mathbb{R}^d$.

### Claim
No, not necessarily.

### Reason
For a conditionally convergent series in $\mathbb{R}^d$, the set of possible sums of convergent rearrangements is governed by the vector version of the Riemann rearrangement phenomenon. In general, this set is not required to be closed under scalar multiplication or addition.

A simple obstruction is that if one rearrangement sum equals $y$, there is no reason that $2y$ must also be attainable by another convergent rearrangement of the same fixed series. Similarly, the sum of two attainable rearrangement limits need not itself be attainable.

So $S$ need not be a linear subspace of $\mathbb{R}^d$.

### A sharper viewpoint
In many cases, the set of rearrangement sums is either:
- a singleton, or
- a translated affine set, or
- much larger but still not linear.

Linear subspace is too strong a requirement here.

Hence the answer is **no**.

### Pitfall alert
A frequent mistake is assuming that because the ambient space is linear, any set of series sums must also be linear. That is false unless you can prove closure under both addition and scalar multiplication. Rearrangement sets usually fail one of those immediately.

### Try different conditions
If the series were **absolutely convergent**, then every rearrangement would converge to the same sum, and the set of rearrangement sums would be a single point. A singleton is a linear subspace only when it is the zero vector, so even in that nicer case the answer is usually still no unless the common sum is $0$.

### Further reading
conditional convergence, rearrangement, linear subspace

Question

Are all rearrangement sums of a conditionally convergent vector series a subspace?

Expert Verified Solution

Step by step

Claim

Reason

A sharper viewpoint

Pitfall alert

Try different conditions

Further reading

FAQ

Is the set of rearrangement sums of a conditionally convergent vector series always a subspace?

What changes if the series is absolutely convergent?