Answer
(a)See the proof: by changing into a telescoping sum, we have:
\[
\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\ldots+\frac{1}{(-1+2 n)(1+2 n)}=\frac{n}{1+2 n}
\]
(b)
\[
\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{(1+2 k)(-1+2 k)}=\frac{1}{2}
\]
Work Step by Step
(a) We will to prove $: \frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\ldots+\frac{1}{(2 n+1)(2 n-1)}=\frac{n}{2 n+1}$
We will change this sum into the telescoping sums by using Partial Fractions for each term:
\[
\frac{1}{(1+2 n)(-1+2 n)}=\frac{1}{2}\left[\frac{1}{2 n-1}-\frac{1}{2 n+1}\right]
\]
Also:
\[
\begin{array}{c}
\frac{1}{1 \cdot 3}=\left(-\frac{1}{3}+1\right)\frac{1}{2} \\
\frac{1}{3 \cdot 5}=\left(-\frac{1}{5}+\frac{1}{3}\right)\frac{1}{2} \\
=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\ldots+\frac{1}{2 n-1}-\frac{1}{2 n+1}\right) \\
=\left(-\frac{1}{1+2 n}+1\right)\frac{1}{2} \\
=\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot .5}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+\ldots+\frac{1}{2}\left(\frac{1}{-1+2 n}-\frac{1}{1+2 n}\right) \\
=\frac{1}{2}\left(\frac{1-1+2 n}{1+2 n}\right) \\
\therefore \frac{1}{1+2 n}
\end{array}
\]
(b) From part (a), we will find:
\[
\begin{aligned}
\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{(1+2 k)(-1+2 k)} &=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\frac{1}{2}\left[\frac{1}{-1+2 k}-\frac{1}{1+2 k}\right]\right) \\
&=\lim _{n \rightarrow \infty} \frac{n}{1+2 n}
\end{aligned}
\]
\[
\begin{array}{l}
=\lim _{n \rightarrow \infty} \frac{n}{1+2 n} \\
=\left[\frac{1}{2}\right]
\end{array}
\]
