jasperE3 wrote: Set $a_n=\frac{2n}{n^4+3n^2+4},n\in\mathbb N$. Prove that $\frac14\le a_1+a_2+\ldots+a_n\le\frac12$ for all $n$. Just write $a_n=b_n-b_{n+1}$ where $b_n=\frac 1{n^2-n+2}$ And so $\sum_{k=1}^na_k=b_1-b_{n+1}=\frac 12-\frac 1{n^2+n+2}$ And problem is just to prove $0\le \frac 1{n^2+n+2}\le\frac 14$ $\forall n\in\mathbb Z_{>0}$, which is immediate