$$\mu = \sum\limits_{k=0}^{\infty}\dfrac{2^{k}}{\displaystyle{2k+1 \choose k}}$$
The answer is $\dfrac{\pi}{2}$. But I don't know how to do it. Please show me the way! Thank you.
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1The answer should be $1+\pi/2$. If $k$ starts at $1$ instead, then you get just $\pi/2$ – BeaumontTaz Sep 14 '14 at 00:54
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@BeaumontTaz Thank you! then, How can i compute this?!? – user175721 Sep 14 '14 at 01:17
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@user170141 What do you mean?!? I can't get it.... – user175721 Sep 14 '14 at 01:21
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Hint: expand the binomial coefficient and do some canceling. – user170141 Sep 14 '14 at 01:23
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@user170141 I did some canceling and the result is. sigma (k+1)!/(2k+1)!! . and from now on, how can I handle this? Thank you again. – user175721 Sep 14 '14 at 01:32
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@user175721, now this question becomes relevant to you. – BeaumontTaz Sep 14 '14 at 01:35
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@BeaumontTaz boom. – user170141 Sep 14 '14 at 01:55