Prove the Taylor series for $f(x)=\sqrt{x+1}$ of $f$ about zero converges to $f$ for all $x$ in $(0,1)$.

Question

This is a multi part question supposed to prove the Taylor series for $f(x)=\sqrt{x+1}$ of $f$ about zero converges to $f$ for all $x$ in $(0,1)$.

part (a) asks:

Let $a_k$=$\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)...(\frac{1}{2}-k+1)}{k!}$ for all $k∈\mathbb{N}$ prove that $\lim_{k\to\infty}$$|\frac{a_k+1}{a_k}|=1$ and use ratio test to conclude that $\sum_{k=0}^{\infty}a_kx^k$ has radius of convergence equal to 1.

I am having a hard time understanding the question, this is how I started:

$a_k$=$\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)...(\frac{1}{2}-(k+1)+1)}{(k+1)!}*\frac{k!}{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)...(\frac{1}{2}-k+1)}$

=$\frac{(\frac{1}{2}-k+2)}{k+1(\frac{1}{2}-k+1)}$

I'm not sure where to go from here.

Any advice would be greatly appreciated.

Thank you,

Answer 1

Since$$a_k=\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\cdots\left(\frac{1}{2}-k+1\right)}{k!},$$you have$$a_{k+1}=\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\cdots\left(\frac{1}{2}-k+1\right)\left(\frac{1}{2}-k\right)}{(k+1)!}=\frac{a_k\left(\frac12-k\right)}{k+1}$$and therefore$$\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty}\left|\frac{\frac12-k}{k+1}\right|=1$$

Answer 2

$$ \left|\frac{a_{k+1}}{a_k}\right|=\left|\frac{\frac{1}{2}-k}{k+1}\right| \underset{k \rightarrow +\infty}{\rightarrow}1 $$