Question

Show that for any \(n \ge 1\)
\[
\frac{1}{2}\le\sum
_{r=1}^n \frac{r}{n^2+r}\le\frac{n+1}{2 n}

\]

Answer 1

\[\sum_{r=1}^{n} \frac{r}{n^2+r} \le \sum_{r=1}^{n} \frac{r}{n^2}=\frac{1}{n^2}\sum_{r=1}^{n}r=\frac{\frac{n(n+1)}{2}}{n^2}=\frac{n+1}{2n}\]

Answer 2

Great. How about the other inequality?

Answer 3

\[\sum_{r=1}^{n} \frac{r}{n^2+r}\ge \sum_{r=1}^{n} \frac{r}{n^2+n}=\frac{1}{2}\]