Test the condition for convergence of ∑n=1∞1n(n+1)(n+2) and find the sum if it exists.

Alternatively, take 11−x=∑n=1∞xn−1 and integrate three times with lower limit 0, giving −log⁡(1−x)=∑n=1∞xnn x+(1−x)log⁡(1−x)=∑n=1∞xnn(n+1) 34x2−12x−12(1−x)2log⁡(1−x)=∑n=1∞xnn(n+1)(n+2)

Step by step solution to "Test the condition for convergence of ∑n=1∞1n(n+1)(n+2) and..."

Sandra Allison

Answered question

2022-01-18

Test the condition for convergence of

\sum_{n = 1}^{\infty} \frac{1}{n (n + 1) (n + 2)}

and find the sum if it exists.

Answer & Explanation

David Clayton

Beginner2022-01-18Added 36 answers

Using Eulers

Karen Robbins

Beginner2022-01-19Added 49 answers

There is an alternate method and is as follows.
Notice that
$\frac{1}{n (n + 1) (n + 2)} = \frac{(n - 1)!}{(n + 2)!} = \frac{12}{B} (n, 3)$
where $B (x, y)$ is the Beta function. Using an integral form of the Beta function the summation becomes
$S = \sum_{n = 1}^{\infty} \frac{1}{n (n + 1) (n + 2}}$
$= \frac{1}{2} \int_{0}^{1} (\sum_{n = 1}^{\infty} x^{n - 1}) (1 - x)^{2} d x$
$= \frac{1}{2} \int_{0}^{1} \frac{(1 - x)^{2}}{1 - x} d x = \frac{1}{2} \int_{0}^{1} (1 - x) d x$
$= \frac{1}{4}$
This leads to the known result
$\sum_{n = 1}^{\infty} \frac{1}{n (n + 1) (n + 2)} = \frac{1}{4}$

alenahelenash

Expert2022-01-24Added 556 answers

Alternatively, take

\frac{1}{1 - x} = \sum_{n = 1}^{\infty} x^{n - 1}

and integrate three times with lower limit 0, giving

- \log (1 - x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n}

x + (1 - x) \log (1 - x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n (n + 1)}

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

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