Verifying Newton's Binomial Series for $(1+x)^{−2}$

Question

Verify the consistency of Newton’s Binomial series for the function $(1+x)^{−2}$ in two ways:

$(a)$ by multiplying the usual geometric series for $(1 +x)^{−1}$with itself

$(b)$ by differentiation of the series for $(1 +x)^{−1}$

$(a)$ I have that Newton's Binomial series says that

$$(1+x)^p=1+px+\frac{p(p−1)}{2!}x^2+\frac{p(p−1)(p−2)}{3!}x^3+···$$

I have that the geometric series for $(1 +x)^{−1}$ is given by

$$\begin{align*} \frac{1}{1+x} &=\sum_{k=0}^{\infty}(-x)^k\\\\ &=-x^0+(-x)^1+(-x)^2+(-x)^3+...\\\\ &=1-x+x^2-x^3+... \end{align*}$$

Thus

$$\begin{align*} \left(\frac{1}{1+x}\right)^2 &=(1-x+x^2-x^3+...)(1-x+x^2-x^3+...)\\\\ &=(1-x+x^2-x^3+...)\\\\ & +(-x+x^2-x^3+...)\\\\ & +(x^2-x^3+...)\\\\ & +(-x^3+...)\\\\ &=1-2x+3x^2-4x^3+... \end{align*}$$

Checking that this equals Newton's Binomial series

$$(1+x)^p=1+px+\frac{p(p−1)}{2!}x^2+\frac{p(p−1)(p−2)}{3!}x^3+···$$

where $p=-2$

$$\begin{align*} (1+x)^{-2} &=1-2x+\frac{-2(-3)}{2!}x^2+\frac{-2(-3)(-4)}{3!}x^3+···\\\\ &=1-2x+3x^2-4x^3+... \checkmark \end{align*}$$

Is this a valid proof?

$(b)$

I have that $$\int_0^x\frac{dt}{1+t}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$$

but I'm not sure how to use this fact.

Answer 1

Yes, first idea is right. More formally, $$ \left(\sum_{i=0}^\infty (-1)^i x^i \right) \left(\sum_{j=0}^\infty (-1)^j x^j \right) = \sum_{n=0}^\infty \sum_{i=0}^n (-1)^i x^i (-1)^{n-i}x^{n-i} = \sum_{n=0}^\infty \sum_{i=0}^n (-1)^n x^n = \sum_{n=0}^\infty (n+1) (-1)^n x^n. $$ On the second one, note that $$ \frac{d}{dx} (1+x)^{-1} = -(1+x)^{-2}, $$ thus $$ (1+x)^{-2} = -\frac{d}{dx} (1+x)^{-1} = -\frac{d}{dx} \sum_{i=0}^\infty (-1)^i x^i = -\sum_{i=0}^\infty (-1)^i \frac{d}{dx} x^i = \sum_{i=1}^\infty (-1)^{i-1}ix^{i-1} = \sum_{i=0}^\infty (-1)^i (i+1)x^i $$

Answer 2

The binomial expansion is $$ \left( {1 + x} \right)^{\, - 2} = \sum\limits_{0\, \le \,k} {\left( \matrix{ - 2 \cr k \cr} \right)x^{\,k} } = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{ k + 1 \cr k \cr} \right)x^{\,k} } = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( {k + 1} \right)x^{\,k} } $$ where "upper negation" of the binomial has been used

The multiplication of geometric series gives $$ \eqalign{ & {1 \over {1 + x}}{1 \over {1 + x}} = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} x^{\,k} } \sum\limits_{0\, \le \,j} {\left( { - 1} \right)^{\,j} x^{\,j} } = \cr & = \sum\limits_{0\, \le \,k} {\sum\limits_{0\, \le \,j} {\left( { - 1} \right)^{\,k + j} x^{\,k + j} } } \quad \left| \matrix{ \;0 \le k + j = s \hfill \cr \;0 \le j = s - k\; \Rightarrow \;0 \le k \le s \hfill \cr} \right.\;\quad = \cr & = \sum\limits_{0\, \le \,s} {\left( {\sum\limits_{0 \le k \le s} 1 } \right)\left( { - 1} \right)^{\,s} x^{\,s} } = \sum\limits_{0\, \le \,s} {\left( { - 1} \right)^{\,s} \left( {s + 1} \right)x^{\,s} } \cr} $$

and differentiating the geometric series $$ \eqalign{ & {d \over {dx}}\left( {1 + x} \right)^{\, - 1} = - \left( {1 + x} \right)^{\, - 2} = {d \over {dx}}\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} x^{\,k} } = \cr & = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} k\,x^{\,k - 1} } = \sum\limits_{1\, \le \,k} {\left( { - 1} \right)^{\,k} k\,x^{\,k - 1} } = \cr & = \sum\limits_{1\, \le \,k} {\left( { - 1} \right)^{\,k + 1} \left( {k + 1} \right)\,x^{\,k} } = - \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( {k + 1} \right)x^{\,k} } \cr} $$