The limit is equal to 0.
Let xn=(2an−a1−a2)(2an−a2−a3)⋯(2an−an−2−an−1)(2an−an−1−a1). The sequence (an)n≥1 is convergent; let L=limn→∞an. Since an≤L for all n≥1 we have 2an−ak−ak+1≤2L−ak−ak+1≤2(L−ak), k=1,2,…,n−2, so 0≤xn≤2n−1(L−a1)(L−a2)⋯(L−an−2)(L−a1)=yn.
Suppose yn>0, otherwise yn=0 and then xn=0. Since ynyn+1=2(L−an)→0, we infer that limn→∞yn=0 and finally limn→∞xn=0.