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The answer is no. E.g., let $x_{2n}=x:=(\frac12,\frac12,\frac13,\frac14,\ldots)$ and $x_{2n+1}=y:=(0,\frac12,\frac13,\frac14,\ldots)$ (for all $n$). Then (i) $x_{2n}=x\to x$ and $x_{2n+1}=y\to y$ in any sense and (ii) $\|x_{2n}+x_{2n+1}\|=2\|x_{2n}\|=2\|x_{2n+1}\|$ and hence $I_n=0$ for all. However, $x\ne y$. $\quad\Box$