Given: \(\displaystyle{E}_{{1}},{E}_{{2}}\ldots{E}{n} ,\) are independent events \(\displaystyle{P}{\left({E}{i}\right)}=\frac{{1}}{{{i}+{1}}}\)

To proof: \(\displaystyle{P}{\left({E}{1}{U}\ldots{U}{E}{n}\right)}=\frac{{n}}{{{n}+{1}}}\)

Use the Complement rule: \(\displaystyle{P}{\left({A}^{{c}}\right)}={P}{\left(\neg{A}\right)}={1}-{P}{\left({A}\right)}\)

\(\displaystyle{P}{\left({\left({E}{i}\right)}^{c}\right)}={1}-{P}{\left({E}{i}\right)}={1}-{\left(\frac{1}{{{i}+{1}}}\right)}=\frac{{{i}+{1}}}{{{i}+{1}}}-{\left(\frac{1}{{{i}+{1}}}\right)}=\frac{i}{{{i}+{1}}}\)

Since the events \(\displaystyle{E}_{{1}},{E}_{{2}}\ldots,{E}_{{n}}\) are independent, the events \(\displaystyle{\left({E}_{1}\right)}^{{c}},{\left({E}_{2}\right)}^{{c}}\ldots{\left({E}_{n}\right)}^{{c}}\) are alslo independent.

We can use the multification rule for independent events \(\displaystyle{P}{\left({A}∪{B}\right)}={P}{\left({A}{\quad\text{and}\quad}{B}\right)}={P}{\left({A}\right)}{x}{P}{\left({B}\right)}\)

\(\displaystyle{P}{\left({\left({E}{1}\right)}^{c}\right)}\cap\ldots \cap{\left({\left({E}{n}\right)}^{c}\right)}={P}{\left({E}{1}\right)}^{c}\cdot\ldots\cdot{P}{\left({\left({E}{n}\right)}^{c}\right)}=\frac{1}{{{1}+{1}}}\cdot{\left(\frac{2}{{{2}+{1}}}\right)}\cdot{\left(\frac{3}{{{3}+{1}}}\right)}\cdot\ldots\cdot{\left(\frac{n}{{{n}+{1}}}=\frac{1}{{2}}\cdot\frac{2}{{3}}\cdot\frac{3}{{4}}\cdot\ldots\cdot\frac{n}{{{n}+{1}}}=\frac{{{1}{\left({2}\right)}{\left({3}\right)}{\left({4}\right)}\ldots{\left({n}\right)}}}{{{2}{\left({3}\right)}{\left({4}\right)}\ldots{\left({n}+{1}\right)}}}=\frac{1}{{{n}+{1}}}\right.}\)

Use the complement rule: \(\displaystyle{P}{\left({A}^{{c}}\right)}={P}{\left(\neg{A}\right)}={1}-{P}{\left({A}\right)}\)

\(P(E_1 \ldots \cup E_n)=P(((E_1)^c)\cap \ldots\cap(E_n)^c)) =1-P(((E_1)^c)\cap \ldots \cap(E_n)^c)=1-\left(\frac{1}{n+1}\right)=\frac{n}{(n+1)}\)