Use binomial series for $\sqrt{1+x}$ to find the binomial series for $\frac{1}{\sqrt{1+x}}$ The question specifically asks me to use the binomial series for $\sqrt{1+x}$ to find the series for $\frac{1}{\sqrt{1+x}}$.
First, I need to find the derivative of the binomial series $\sqrt {1+x}$
I first constructed the binomial series based on the formula $$1+rx+\frac {r(r-1)x^2}{2!} + ... + \frac{r(r-1)...(r-n+1)x^n}{n!}$$
I got $$1+\frac {1}{2} x - \frac{1}{2!}\frac{1}{2^{2}}x^2 + ... + \frac{(-1)^{n+1}}{n!}\frac{1\cdot 3\cdot 5...(2n-3)}{2^n}x^n$$
$$1 + \Sigma_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}\frac{1\cdot 3 \cdot 5...\cdot (2n-3)}{2^n}x^n$$
When I take the derivative, the first few terms becomes:
$$0 + \frac{1}{2} - \frac{1}{1!}\frac{1}{2^{2}}x + \frac{1}{2!}\frac{1\cdot3}{2^3}x^2 ... $$
So $$\frac{d}{dx}\sqrt{1+x} \,=\, \frac{1}{2} + \Sigma_{n=1}^{\infty} \frac{(-1)^n}{n!} \frac{1\cdot 3 \cdot 5...\cdot (2n-3)}{2^{n+1}}x^{n}$$
And I also know that $$\frac{d}{dx}\sqrt{1+x} = \frac{1}{2\sqrt{1+x}}$$
Therefore, $$\frac{1}{\sqrt{1+x}} = 2\frac{d}{dx} \sqrt{1+x}$$
So multiply the series I found by 2 gives me $$2\cdot (\frac{1}{2} + \Sigma_{n=1}^{\infty} \frac{(-1)^n}{n!} \frac{1\cdot 3 \cdot 5...\cdot (2n-3)}{2^{n+1}}x^{n}) = 1 + \Sigma_{n=1}^{\infty} \frac{(-1)^n}{n!} \frac{1\cdot 3 \cdot 5...\cdot (2n-3)}{2^{n}}x^{n}$$
However, my answer was wrong, and instead of $$1\cdot 3 \cdot 5...\cdot (2n-3)$$, it should be $$1\cdot 3 \cdot 5...\cdot (2n-1)$$
Why is that?

The expanded form of $\frac{d}{dx}\sqrt{1+x}$ is correct, the compact one is not. You could do $$\frac{d}{dx}\left(1+\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n!}\frac{1\cdot3\cdots(2n-3)}{2^n}x^n\right)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(n-1)!}\frac{1\cdot3\cdots(2n-3)}{2^n}x^{n-1}$$ without expanding (at $n=1$ we have an empty product in both cases, which is assumed to be $1$).
Now, to "shift the index", put $n=m+1$ everywhere, and you get $$\sum_{m=\color{red}{0}}^\infty\frac{(-1)^m}{m!}\frac{1\cdot3\cdots(2m-1)}{2^{m+1}}x^m.$$
