Answer
See the full explanation below.
Work Step by Step
The expression is given below:
${{S}_{n}}=\left( 1-\frac{1}{2} \right)\times \left( 1-\frac{1}{3} \right)\times \left( 1-\frac{1}{4} \right)........\left( 1-\frac{1}{\left( n+1 \right)} \right)$
And the above sum can be written in a general from
${{S}_{n}}=\prod\limits_{n=1}^{n}{\left( 1-\frac{1}{(n+1)} \right)}$
It can be simplified as
$\begin{align}
& {{S}_{n}}=\prod\limits_{n=1}^{n}{\left( 1-\frac{1}{(n+1)} \right)} \\
& {{S}_{n}}=\prod\limits_{n=1}^{n}{\left( \frac{n}{n+1} \right)} \\
\end{align}$
Substitute the values of n in the above equation
${{S}_{n}}=\left( \frac{1}{2} \right)\times \left( \frac{2}{3} \right)\times \left( \frac{3}{4} \right)......\left( \frac{n}{\left( n+1 \right)} \right)$
${{S}_{n}}=\frac{1}{\left( n+1 \right)}$ $(I)$
To prove the conjectural formula by mathematical induction,
Let ${{S}_{n}}$ be a statement involving the positive integer n. If
1. ${{S}_{1}}$ is true, and
2. Let us assume that the ${{S}_{k}}$ be true for some integer $ k $.
Then, one has to use ${{S}_{k}}$ to prove ${{S}_{k+1}}$ is also true. If one is able to prove ${{S}_{k+1}}$ then by the principle of mathematical induction the statement ${{S}_{n}}$ is true for all positive integers $ n $.
For ${{S}_{1}}$ put $ n=1$ in the above equation, one gets
${{S}_{1}}=\frac{1}{2}$,
It is the same as the first term; thus the first equation is satisfied.
Then, let us assume that ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ should also be true by the method of mathematical induction.
${{S}_{k}}=\frac{1}{\left( k+1 \right)}$ $(II)$
Then, one has to prove ${{S}_{k+1}}$ is true
${{S}_{k+1}}=\frac{1}{k+2}$
Now, one has
$\begin{align}
& {{S}_{k+1}}={{S}_{k}}\times \left( \frac{k+1}{k+2} \right) \\
& {{S}_{k+1}}=\left( \frac{1}{k+1} \right)\times \left( \frac{k+1}{k+2} \right) \\
\end{align}$
And simplify
${{S}_{k+1}}=\frac{1}{k+2}$
This inequality for $ k+1$ is the same as equation (I); this means it has satisfied the second condition of mathematical induction.
So, the expression ${{S}_{n}}=\left( 1-\frac{1}{2} \right)\times \left( 1-\frac{1}{3} \right)\times \left( 1-\frac{1}{4} \right)........\left( 1-\frac{1}{\left( n+1 \right)} \right)$ is true for all positive integers $ n $.
Thus, the value are ${{S}_{1}}=\frac{1}{2}$, ${{S}_{2}}=\frac{1}{3}$, ${{S}_{3}}=\frac{1}{4}$, ${{S}_{4}}=\frac{1}{5}$ and ${{S}_{5}}=\frac{1}{6}$.
