What's the Final Number ?

First notice that as terms $a, b, c,...$ are successively combined, we end up with $a + b + c +...+ ab + bc + ca + .....abc...$ , so that the final answer is independent of what order the numbers are combined, i.e. symmetric with all the terms $a, b, c,...$ . Next, notice that $n$ and $1/(n+1)$ , combined, results in $n+1$ . Hence, $1$ and $1/2$ results in $2$ , $2$ and $1/3$ results in $3$ , etc. The final term after all in S are combined is $100$ .

Conjecture: If the process is repeated on a sequence $\left\{ { a }_{ 1 },{ a }_{ 2 },...{ ,a }_{ n } \right\}$ , the result is $({ a }_{ 1 }+1)({ a }_{ 2 }+1)...({ a }_{ n }+1)-1$ .

We proof by induction.

Clearly $n=1$ works.

Assume $n=k$ works.

Then we show $n=k+1$ works.

Repeating the process on the $\left\{ { a }_{ 1 },{ a }_{ 2 },...{ ,a }_{ k },{ a }_{ k+1 } \right\}$ k times. WLOG, we let the process be repeated on first k terms. Then, the set becomes $\{ ({ a }_{ 1 }+1)({ a }_{ 2 }+1)...({ a }_{ k }+1)-1,{ a }_{ k+1 }\}$ . Repeating the process once more, we get $({ a }_{ 1 }+1)({ a }_{ 2 }+1)...({ a }_{ k }+1)-1+{ a }_{ k+1 }+{ a }_{ k+1 }[({ a }_{ 1 }+1)({ a }_{ 2 }+1)...({ a }_{ k }+1)-1]\\ ={ a }_{ k+1 }({ a }_{ 1 }+1)({ a }_{ 2 }+1)...({ a }_{ k }+1)+({ a }_{ 1 }+1)({ a }_{ 2 }+1)...({ a }_{ k }+1)-1\\ ={ (a }_{ k+1 }+1)({ a }_{ 1 }+1)({ a }_{ 2 }+1)...({ a }_{ k }+1)-1$

Therefore, by induction, our conjecture is true, and the final number is $2\quad \times \quad \frac { 3 }{ 2 } \quad \times \quad \frac { 4 }{ 3 } \quad \times \quad ...\quad \times \quad \frac { 101 }{ 100 } \quad -\quad 1\quad = \boxed {100}$

I also considered the pattern of the remained number, similarly with the solution from Michael Mendrin.

Notice the pattern of the remained number.

Consider the polynomial $f(x)=(x-1)\left(x-\frac{1}{2}\right)\left(x-\frac{1}{3}\right)\cdots\left(x-\frac{1}{100}\right).$ By Vieta's formula, the remained number is exactly the value of $f(-1)-(-1)^{100}$ and we have $f(-1)-(-1)^{100} = -2\left(-\frac{3}{2}\right)\left(-\frac{4}{3}\right)\cdots\left(-\frac{101}{100}\right)-1=\boxed{100}.$