I'm trying to prove the Taylor expansion of $(1+x)^\alpha$ when $\alpha$ is not positive integers.
However I stuck at the point of proving of the residual of Taylor expansion converges to ZERO.
If I define $\binom \alpha n = \frac{\alpha(\alpha-1)(\alpha-2)...(\alpha -n+1)}{n!}$ the residual of Taylor expansion at $x=0$ is:
\begin{aligned} &R_n(x)=\lim_{n\to\infty}\left\{f(x)-\sum_{i=0}^n\frac{f^{(i)}(x_0)}{i!}(x-x_0)^n\right\} \\ &=\lim_{n\to\infty}\left\{ (1+x)^\alpha - \sum_{n=0}^\infty \binom \alpha n x^n\right\} \end{aligned}
And given that $x=1,-1<\alpha<0$ , the residual should be converging to ZERO (according to by book, which didn't give out prove for that), hence the equation I wanna prove is:
\begin{aligned} &R_n(1)=\lim_{n\to\infty}\left\{ (1+1)^\alpha - \sum_{n=0}^\infty \binom \alpha n\right\}=0 \end{aligned}
Giving the condition that $-1<\alpha<0$ .
If anyone can help me on proving that or, prove it is not correct. Thanks!!
The integral form of the remainder for $x=1$ is $$ (\alpha-n) \binom{\alpha}{n} \int_0^1 (1-u)^n (1+u)^{\alpha-n-1} \, du = (\alpha-n) \binom{\alpha}{n} \int_0^1 \left( \frac{1-u}{1+u} \right)^n (1+u)^{\alpha-1} \, du. $$ On the interval of integration $(1-u)/(1+u)<1-u$, so the integral is bounded by $$ \int_0^1 (1-u)^n 2^{\alpha-1} \, du = \frac{2^{\alpha-1}}{n+1}. $$ So the remainder is bounded above by $$ 2^{\alpha-1}\left| \frac{\alpha-n}{n+1}\binom{\alpha}{n} \right| = 2^{\alpha-1}\left| \binom{\alpha}{n+1} \right|. $$ Can you show that this tends to zero? Stirling's formula's probably the easiest way, but it should also be possible to look at quotients and find that the ratio is smaller than 1.