Answer
(a) By changing into the telescoping sums, we have:
\[
\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{(1+n)n}=\frac{n}{1+n}
\]
(b) $\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{1}{(1+k)k}=1$
Work Step by Step
(a) We will \prove $=\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{n(n+1)}=\frac{1+n}{n}$
By using Partial Fraction for each term we will change this sum into the telescoping sums :
\[
\frac{1}{(1+n)n}=-\frac{1}{1+n}+\frac{1}{n}
\]
Also:
\[
\begin{array}{l}
\qquad \begin{array}{l}
\frac{1}{1 \cdot 3}=-\frac{1}{2}+1 \\
\frac{1}{3 \cdot 5}=-\frac{1}{3}+\frac{1}{2} \\
\frac{1}{3 \cdot 5}=-\frac{1}{4}+\frac{1}{3}
\end{array} \\
\begin{array}{l}
=\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{2}-\frac{1}{4} \dots+\frac{1}{n}-\frac{1}{1+n}\right) \\
=-\frac{1}{1+n}+1 \\
\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{(1+n)n}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{1}+\left(\frac{1}{2}-\frac{1}{3}\right)\right) \dots+\left(\frac{1}{n}-\frac{1}{1+n}\right)
\end{array} \\
=\left(\frac{n+1-1}{n+1}\right)\frac{1}{2}
\end{array}
\]
$=\left[\frac{n}{1+n}\right]$
\[
\frac{1}{2 \cdot 3}+\frac{1}{1 \cdot 2}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{(1+n)n}=\frac{n}{1+n}
\]
$(b)$
\[
\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{1}{(1+k)k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n} -\frac{1}{1+k}+\frac{1}{k}
\]
\[
\begin{array}{l}
=\lim _{n \rightarrow+\infty} \frac{n}{1+n} \\
=1
\end{array}
\]
