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I came across the the function $\ln(1+\sqrt{1-x})$ when I tried to evaluate the sum $\sum_{n=1}^\infty \frac{c^n}{n} \binom{2n}{n}$ for some small c.

Specifically, Mathematica noted that $$ \ln(1+\sqrt{1-x}) = \ln{2} - \frac{1}{2}\sum_{n=1}^\infty \frac{1}{n} \Big(\frac{x}{4}\Big)^n \binom{2n}{n} .$$

How should I show this? So far I have tried expanding $\ln(1+\sqrt{1-x})$ in powers of $\sqrt{1-x}$ and then expanding each of those powers in turn, but I'm having difficulty handling the resulting sum. Taking the nth derivative of the function at left is also proving difficult.

Dylan
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user196574
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  • A good demonstration of the power of mathematica (+1) – Peter May 27 '18 at 09:17
  • What happens when you differentiate it? – Akiva Weinberger May 27 '18 at 09:17
  • I had been trying to differentiate it a few times in a row and figure out some inductive step, but I found the number of terms was growing at a decent pace. As Lord Shark answered below, a single derivative and a small manipulation were all it took to bring the square root into the numerator, which helped me quite a bit. – user196574 May 27 '18 at 09:35

1 Answers1

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$$\frac d{dx}\ln(1+\sqrt{1-x})=-\frac12\frac{(1-x)^{-1/2}}{1+\sqrt{1-x}} =-\frac12\frac{(1-x)^{-1/2}(1-\sqrt{1-x})}x =\frac{1-(1-x)^{-1/2}}{2x}.$$ Now expand by the binomial theorem, and integrate.

Angina Seng
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