"Is there a closed-form of∑n=0∞(−1)n(2n+1)(2n+2)(2n+3)(2n+4)" was answered by our community

trasbocohf1

Answered question

2022-02-22

Is there a closed-form of

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1) (2 n + 2) (2 n + 3) (2 n + 4)}

blokova5u8

Beginner2022-02-23Added 7 answers

Let we sketch an alternative technique. Since:

\sum_{n \geq 0} x^{2 n} = \frac{1}{1 - x^{2}} = \frac{1}{2} (\frac{1}{1 + x} + \frac{1}{1 - x})

we have:

\sum_{n \geq 0} \frac{x^{2 n + 1}}{2 n + 1} = \frac{1}{2} (\log (1 + x) - \log (1 - x)) = \arctan x

and integrating again:

\sum_{n \geq 0} \frac{x^{2 n + 1}}{(2 n + 1) (2 n + 2)} = x \arctan x + \frac{1}{2} \log (1 - x^{2})

and by setting

g (x) = \sum_{n \geq 0} \frac{x^{2 n + 4}}{(2 n + 1) (2 n + 2) (2 n + 3) (2 n + 4)}

we have:

g (x) = \frac{1}{12} (- 5 x^{2} + (2 x^{2} + 6 x) \frac{\arctan \sqrt{x}}{\sqrt{x}} + (3 x + 1) \log (1 - x))

and finally:

\sum_{n \geq 0} \frac{{(- 1)}^{n}}{(2 n + 1) (2 n + 2) (2 n + 3) (2 n + 4)} = \frac{5 - π - 2 \log 2}{12}

emeriinb4r

Beginner2022-02-24Added 10 answers

To give another approach:
The sum can be calculated using the observation that

\int_{0}^{1} {(1 - x^{\frac{1}{n}})}^{k} d x = \frac{1}{(\begin{matrix} n + k \\ n \end{matrix})}

which is related to the beta function. Using this we get:

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1) (2 n + 2) (2 n + 3) (2 n + 4)} = \frac{1}{4!} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 4), (4)} = \frac{1}{4!} \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} {(1 - x^{\frac{1}{4}})}^{2 n} d x

= \frac{1}{4!} \int_{0}^{1} \sum_{n = 0}^{\infty} {(- 1)}^{n} {(1 - x^{\frac{1}{4}})}^{2 n} d x

= \frac{1}{3!} \int_{0}^{1} \frac{1 - 3 x + 3 x^{2} - x^{3}}{1 + x^{2}} d x

= \frac{1}{6} (- \frac{π}{2} - \ln (2) + \frac{π}{2})

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