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?
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1yes I think it will work. – Emptymind Jun 07 '17 at 05:33
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WolframAlpha says the ratio test is inconclusive. It exceeds standard computation time if I try and check it is equal to $0$ – lioness99a Jun 07 '17 at 09:41
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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 $-1+\sqrt{1-t}$.
Angina Seng
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@LordSharktheUnknown Can you explicitly use the Stirling formula to justify the order of $1/k^{3/2}$? – Jun 07 '17 at 07:19
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@Elliot To use Stirling, you express $\binom{1/2}k$ in terms of factorials and the expression $1\times 3\times 5\times\cdots (2k-3)$. But that is $(2k-2)!/((k-1)!2^k)$ so that's can be expressed by factorials too. – Angina Seng Jun 07 '17 at 18:31
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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 $n,x?$
lab bhattacharjee
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