Expert Answer to "How does one sum the given series:12n+1+12⋅12n+3+1⋅32⋅412n+5+1⋅3⋅52⋅4⋅612n+7+…"

Ali Marshall

Answered question

2022-04-23

How does one sum the given series:

\frac{1}{2 n + 1} + \frac{1}{2} \cdot \frac{1}{2 n + 3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{2 n + 5} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{1}{2 n + 7} + \dots

Killian Curry

Beginner2022-04-24Added 18 answers

Observe that

\frac{1}{4^{n}} (\begin{matrix} 2 n \\ n \end{matrix}) = \frac{(2 n - 1) (2 n - 3) (2 n - 5) \dots}{2 n (2 n - 2) (2 n - 4) \dots}

and recall that

\frac{1}{\sqrt{1 - x^{2}}} = \sum_{k \geq 0} \frac{1}{4^{k}} (\begin{matrix} 2 k \\ k \end{matrix}) x^{2 k}

It follows that the desired quantity is

\int_{0}^{1} \frac{x^{2 n}}{\sqrt{1 - x^{2}}} d x

But letting

x = \sin θ

this is just

\int_{0}^{\frac{π}{2}} \sin^{2 n} θ d θ = \frac{π}{2} \frac{1}{4^{n}} (\begin{matrix} 2 n \\ n \end{matrix})

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