First, we will prove that $0 \leq a_{k}-a_{k+1}$ for all $k\geq 1$. If not, let $ a_{k}-a_{k+1}=d<0$ for some $k\geq 1$. As we are given $ a_{k}-a_{k+1}\geq a_{k+1}-a_{k+2}$, it follows $a_{k+i-1}-a_{k+i}\le d<0$ for all $i>0$. Moreover, by summing them all, $a_{k}-a_{k+i}\le id$. For large enough $i$, we must have $a_{k+i}<0$, which contradicts the fact the sequence is made of non-negative real numbers. Therefore $0 \leq a_{k}-a_{k+1}$ for all $k\geq 1$. Second, we will prove that $a_{k}-a_{k+1} <\frac{2}{k^{2}}$. We have $1\ge \sum_{i=1}^{k}a_i\ge \sum_{i=1}^{k}a_i-ka_{k+1}= \sum_{i=1}^{k} i(a_{i}-a_{i+1}) \ge \frac{k(k+1)}{2}(a_{k}-a_{k+1})$. Therefore $a_{k}-a_{k+1}\le \frac{2}{k(k+1)} <\frac{2}{k^{2}}$.

The argumentation in the first part is wrong. From $a_{k}-a_{k+i}\le id < 0$, even for large enough $i$, it does not follow $a_{k+i}<0$. A simple example is for $a_n=n$ for all $n\geq 1$, when $d=-1<0$ for all $n$. In order to obtain that result, we need also use the second condition, namely $1\ge \sum_{i=1}^{k}a_i$ for all $k$. As above, if $a_{k}-a_{k+1} = d < 0$ for some $k$, it follows $a_{k}-a_{k+i}\le id < 0$ for all $i$; but then $a_{k+i} \geq a_k - id \geq i|d|$, and so $\lim_{n\to \infty} a_n = +\infty$, absurd.