How to find the sum of this series? $$\sum_{k=0}^{\infty}\cfrac{{2}^{k}}{\binom{2k+1}{k}}$$
It seems very easy. But I still can not work it out, can anyone help?
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2"it seems very easy." Why do you say so? – Did Sep 12 '14 at 18:05
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Check this technique. – Mhenni Benghorbal Sep 13 '14 at 21:32
2 Answers
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We have, using the Euler Beta function: $$\begin{eqnarray*}\color{red}{\sum_{k=0}^{+\infty}\frac{2^k}{\binom{2k+1}{k}}}&=&\sum_{k=0}^{+\infty}\frac{2^k \Gamma(k+1)\Gamma(k+2)}{\Gamma(2k+2)}=\sum_{k=0}^{+\infty}2^k(k+1)\,B(k+1,k+1)\\&=&\sum_{k=0}^{+\infty}2^k(k+1)\int_{0}^{1}x^k(1-x)^k\,dx\\&=&\int_{0}^{1}\frac{dx}{(1-2x(1-x))^2}=\color{red}{\frac{\pi}{2}+1}.\end{eqnarray*}$$
Jack D'Aurizio
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1Based on that final result, I glanced at Mathworld's page for Pi formulas and found that it appears as equation 23. A few lines later it points out that it can be derived by expanding $\dfrac{\sin^{-1}x}{\sqrt{1-x^2}}$ in powers of $x$ and plugging in $x=1/\sqrt{2}$. – Semiclassical Sep 12 '14 at 19:09
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This answer is similar to Eric Naslund's answer to this question. (+1) – robjohn Sep 13 '14 at 00:59
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@robjohn: Thanks for pointing that out. Central binomials in the denominators often arise from the Euler acceleration technique; my method is the same as Eric Naslund's, that's true. – Jack D'Aurizio Sep 13 '14 at 01:05
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1Thank you!It's really amazing,I never consider to use Euler Beta function to solve this question,it seems really easy now. – Hex Sep 13 '14 at 02:08
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This question is closely related to this question.
$$
\begin{align}
\sum_{k=0}^\infty\frac{2^k}{\binom{2k+1}{k}}
&=\sum_{k=0}^\infty2^k\frac{k!(k+1)!}{(2k+1)!}\tag{1}\\
&=\sum_{k=0}^\infty\frac{(k+1)!}{(2k+1)!!}\tag{2}\\
&=\sum_{k=0}^\infty\frac{k!}{(2k+1)!!}
+\sum_{k=0}^\infty\frac{k\,k!}{(2k+1)!!}\tag{3}\\
&=\sum_{k=0}^\infty\frac{k!}{(2k+1)!!}
+\sum_{k=0}^\infty\left(\frac{k!}{(2k-1)!!}-\frac{(k+1)!}{(2k+1)!!}\right)\tag{4}\\
&=\frac\pi2+1\tag{5}
\end{align}
$$
Explanation:
$(1)$: rewrite binomial coefficient with factorials
$(2)$: $(2k+1)!!=\frac{(2k+1)!}{2^kk!}$
$(3)$: $(k+1)!=k!+k\,k!$
$(4)$: $\frac{k!}{(2k-1)!!}-\frac{(k+1)!}{(2k+1)!!}=\frac{(2k+1)k!}{(2k+1)!!}-\frac{(k+1)k!}{(2k+1)!!}=\frac{k\,k!}{(2k+1)!!}$
$(5)$: this answer and telescoping series
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(+1) Your answers are always admirable and insightful, I am really glad you are in the "leadership" of this community. – Jack D'Aurizio Sep 13 '14 at 01:08