How to find the sum of this series? ∑k=0∞2k(2k+1k)

We have, using the Euler Beta function: ∑k=0+∞2k(2k+1k)=∑k=0+∞2kΓ(k+1)Γ(k+2)Γ(2k+2) =∑k=0+∞2k(k+1)∫01xk(1−x)kdx =∫01dx(1−2x(1−x))2=π2+1

Rewrite binomial coefficient with factorials ∑k=0∞2k(2k+1k)=∑k=0∞2kk!(k+1)!(2k+1)! =∑k=0∞(k+1)!(2k+1)! =∑k=0∞k!(2k+1)!!+∑k=0∞kk!(2k+1)!! =∑k=0∞k!(2k+1)+∑k=0∞(k!(2k−1)!!−(k+1)!(2k+1)!!) =π2+1

We have, using the Euler Beta function: ∑k=0+∞2k(2k+1k)=∑k=0+∞2kΓ(k+1)Γ(k+2)Γ(2k+2) =∑k=0+∞2k(k+1)∫01xk(1−x)kdx =∫01dx(1−2x(1−x))2=π2+1

Rewrite binomial coefficient with factorials ∑k=0∞2k(2k+1k)=∑k=0∞2kk!(k+1)!(2k+1)! =∑k=0∞(k+1)!(2k+1)! =∑k=0∞k!(2k+1)!!+∑k=0∞kk!(2k+1)!! =∑k=0∞k!(2k+1)+∑k=0∞(k!(2k−1)!!−(k+1)!(2k+1)!!) =π2+1

Best solutions to - "How to find the sum of this series?..."

Kasey Haley

Answered question

2022-01-22

How to find the sum of this series?

\sum_{k = 0}^{\infty} \frac{2^{k}}{(\begin{matrix} 2 k + 1 \\ k \end{matrix})}

Answer & Explanation

trasahed

Beginner2022-01-23Added 14 answers

We have, using the Euler Beta function:

\sum_{k = 0}^{+ \infty} \frac{2^{k}}{(\begin{matrix} 2 k + 1 \\ k \end{matrix})} = \sum_{k = 0}^{+ \infty} \frac{2^{k} Γ (k + 1) Γ (k + 2)}{Γ (2 k + 2)}

= \sum_{k = 0}^{+ \infty} 2^{k} (k + 1) \int_{0}^{1} x^{k} {(1 - x)}^{k} d x

= \int_{0}^{1} \frac{d x}{{(1 - 2 x (1 - x))}^{2}} = \frac{π}{2} + 1

trovabile4p

Beginner2022-01-24Added 13 answers

Rewrite binomial coefficient with factorials

\sum_{k = 0}^{\infty} \frac{2^{k}}{(\begin{matrix} 2 k + 1 \\ k \end{matrix})} = \sum_{k = 0}^{\infty} 2^{k} \frac{k! (k + 1)!}{(2 k + 1)!}

= \sum_{k = 0}^{\infty} \frac{(k + 1)!}{(2 k + 1)!}

= \sum_{k = 0}^{\infty} \frac{k!}{(2 k + 1)!!} + \sum_{k = 0}^{\infty} \frac{k k!}{(2 k + 1)!!}

= \sum_{k = 0}^{\infty} \frac{k!}{(2 k + 1)} + \sum_{k = 0}^{\infty} (\frac{k!}{(2 k - 1)!!} - \frac{(k + 1)!}{(2 k + 1)!!})

= \frac{π}{2} + 1

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