$\operatorname{sconv}(A)=\left\{ \sum_{i=1}^\infty\lambda_i\cdot a_i:a_i\in A, \lambda_i\ge0,\sum_{i=1}^\infty\lambda_i=1\right\}$ is superconvex

Question

Let $X$ be a Banach space and $A\subset X$ a subset bounded.

Denote by $\operatorname{sconv}(A)$ the superconvex hull of $A$:

$$\operatorname{sconv}(A)=\left\{ \sum_{i=1}^{\infty}\lambda_i\cdot a_i:\;\;a_i\in A\;,\;\lambda_i\ge0\;,\;\sum_{i=1}^\infty \lambda_i=1\right\}$$

A set $A$ is superconvex if $\operatorname{sconv}(A)=A$

How can we prove that: $\operatorname{sconv}(A)$ is superconvex

Any hints would be appreciated.