which proves the Binomial Theorem for any positive fractional index.
534. To prove the Binomial Theorem when the index is any negative quantity.
It has been proved that
f(m) \times f(n) = f(m + n)
for all values of m and n. Replacing m by -n (where n is positive), we have
f(-n) \times f(n) = f(-n + n) = f(0) = 1
since all terms of the series except the first vanish;
\frac{1}{(f(n)} = f(-n)
but f(n)=(1 + x)^n for any positive value of n;
\frac{1}{(1 + x)^n} = f(-n)
or (1 + x)^{-n} = f(-n).
But f(-n) stands for the series
which proves the Binomial Theorem for any negative index.
535. It should be noticed that when x<1, each of the series f(m), f(n), f(m+n) is convergent, and f(m +n) is the true arithmetical equivalent of f(m) \times f(n). But when x>1, all these series are divergent, and we can only assert that if we multiply the series denoted by f(m), by the series denoted by f(n), the first r terms of the product will agree with the first r terms of f(m+n), whatever finite value r may have.
