Problem 1116

Solution. Let's represent each term $\frac{k}{n^{2}+k}$ on the left side of the inequality as a difference $\frac{k}{n^{2}}-\frac{k^{2}}{n^{2}\left(n^{2}+k\right)}$ and add the resulting equalities term by term:

$$\begin{aligned} & \frac{1}{n^{2}+1}+\frac{2}{n^{2}+2}+\cdots+\frac{n}{n^{2}+n} \\ & \quad=\frac{1+2+\cdots+n}{n^{2}}-\left(\frac{1}{n^{2}\left(n^{2}+1\right)}+\frac{4}{n^{2}\left(n^{2}+2\right)}+\cdots+\frac{n^{2}}{n^{2}\left(n^{2}+n\right)}\right) \end{aligned}$$

Now, let's replace the denominators of all terms in the parentheses with $n^{2}\left(n^{2}+n\right)$, which will only increase the sum and make it equal to

$$\begin{aligned} \frac{1+2+\cdots+n}{n^{2}} & -\frac{1+4+\cdots+n^{2}}{n^{2}\left(n^{2}+n\right)}=\frac{n(n+1)}{2 n^{2}}-\frac{n(n+1)(2 n+1)}{6 n^{3}(n+1)}=\frac{n(n+1)}{2 n^{2}}-\frac{(2 n+1)}{6 n^{2}} \\ = & \frac{3 n^{2}+n-1}{6 n^{2}}=\frac{1}{2}+\frac{n-1}{6 n^{2}}<\frac{1}{2}+\frac{1}{6 n} \end{aligned}$$

Comment. Correct solution - 20 points. Obtained a close estimate of the sum - 10 points. There is a gap in the justification - 15 points. There is some progress in the solution - 5 points. The solution is started, but the progress is insignificant - 1 point. The solution is incorrect or missing - 0 points.