Let for some $k$ we have $a_k-a_{k+1}<0$ Then $a_{k+2}-a_{k+1} \geq a_{k+1}-a_k>0 \to a_{k+i+1}>a_{k+i}>a_k$ for natural $i$ $\sum_{j=1}^{k+n} a_j=\sum_{j=1}^{k-1} a_j+\sum_{j=k}^{k+n} a_j> \sum_{j=1}^{k-1} a_j+(n+1)a_k>1$ for some $n$, so we get contradiction. So $a_k-a_{k+1} \geq 0$ Let $a_k-a_{k+1} \geq \frac{2}{k^2}$ Then $a_1-a_2 \geq a_2-a_3 \geq ... \geq a_{k-1}-a_k\geq a_k-a_{k+1} \geq \frac{2}{k^2}$ For $n<k$ we have $a_n=(a_n-a_{n+1})+(a_{n+1}-a_{n+2})+...+(a_k-a_{k+1})+a_{k+1} \geq \frac{2(k+1-n)}{k^2} $ $\sum_{j=1}^{k} a_j \geq \sum_{j=1}^{k} \frac{2(k+1-j)}{k^2}=\frac{2(k+1)}{k}- \sum_{j=1}^{k} \frac{2j}{k^2}=2+\frac{2}{k}-\frac{k+1}{k}=\frac{k+1}{k}>1$ so we get contradiction