(a) We start with \(\sum_{r=1}^{n} (2r+1) = 2\sum_{r=1}^{n} r + n\). Using the formula for the sum of the first \(n\) natural numbers, \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\), we have:

\(2\sum_{r=1}^{n} r + n = 2 \times \frac{n(n+1)}{2} + n = n(n+1) + n = n(n+2)\).

(b) To show \(\frac{2r+1}{(r^2+1)(r^2+2r+2)} = \frac{1}{r^2+1} - \frac{1}{r^2+2r+2}\), we find a common denominator:

\(\frac{1}{r^2+1} - \frac{1}{r^2+2r+2} = \frac{r^2+2r+2 - (r^2+1)}{(r^2+1)(r^2+2r+2)} = \frac{2r+1}{(r^2+1)(r^2+2r+2)}\).

(c) Using the result from (b), the series becomes a telescoping series:

\(\sum_{r=1}^{n} \left( \frac{1}{r^2+1} - \frac{1}{r^2+2r+2} \right) = \frac{1}{2} - \frac{1}{n^2+2n+2}\).

(d) As \(n \to \infty\), the term \(\frac{1}{n^2+2n+2} \to 0\), so the sum converges to \(\frac{1}{2}\).