Uniform convergence of the series for $\frac{x}{\sqrt{1+x}} = \sum_{n=1}^\infty (-1)^n\binom{-1/2}{n}\left(\frac{x}{1+x}\right)^{n+1}$ I am looking for the values where this series expansion converges uniformly.
$$\frac{x}{\sqrt{1+x}} = \sum_{n=1}^\infty (-1)^n\binom{-1/2}{n}\left(\frac{x}{1+x}\right)^{n+1}$$
Intuitively, I believe the answer is $|x| < 1$, but I am not sure how I can show that using the Weierstrass Majorant test or the Abel test. I'm guessing some sort of manipulation is required so that I can easily use Abel's test would be the best way to solve, but I am not sure on how to attack it.
 A: The solution by Igor Rivin is not correct.  We are interested in uniform convergence of this series for values of x.  The question is valid. You can easily see show with the ratio test that the series is absolutely convergent.  you will get the condition stated by Igor above
$$
|x| < \big|1+x\big|.
$$
Thus we can see $x > -1/2 $ satisfies this condition.  The series will be uniformly convergent for $x > -1/2$ which can be shown using Weierstrass M-test with $M=x/(x+1)$. 
NOW, we have to check for uniform convergence on the endpoint of the interval $x=-1/2$.  At this value we can see the series is alternating and conditionally convergent.  We showed it was not absolutely convergent by the ratio test, however it is conditionally convergent by the alternating series test.  This consists of showing the series term $a_n$ vanishes in the $\lim_{n \to \infty}$ and showing that it is monotonically decreasing.  
Since the series is conditionally convergent at $x=-1/2$, the Weierstrass M-Test fails.  We are forced to use another method,now you are ready to use the Supremum to show that the series is Uniformly Convergent for $x=[-1/2+\epsilon, N]$ where $\epsilon > 0$ and small, N is large.  Note, $\infty$ is not included in the uniform convergence interval. 
A: The binomial coefficient is asymptotic $n^{-1/2},$ which means that the series converges absolutely when $|x| < |1+x|.$
