TITLE: Show this infinite series converges
QUESTION [5 upvotes]: I want to show that the series $$\sum_{k=1}^{\infty} \binom{\frac{1}{2}}{k} (-1)^k$$ converges. I'm fairly sure it converges to zero, but haven't been successful. Will the ratio test work?

REPLY [4 votes]: The ratio test will fail. But the series is absolutely
convergent. The $k$-th term
is
$$-\frac{(1/2)(1/2)(3/2)\cdots(k-3/2)}{k!}.$$
This is $O(1/k^{3/2})$ by say Stirling formula, or that its logarithm is
a constant plus $\sum_{j=2}^k\ln(1-3/(2k))$.
By Abel's theorem, The sum is
$$\lim_{t\to 1^-}\sum_{k=1}^\infty\binom{1/2}k(-1)^kt^k$$
which is $-1$ since this function is $-1+\sqrt{1-t}$.

REPLY [2 votes]: HINT:
Like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $ using 
 Binomial series,
$$nx=\dfrac12(-1)$$
$$\dfrac{n(n-1)x^2}2=\dfrac{1/2(1/2-1)}{2!}(-1)^2$$
Can you solve for $n,x?$