Key to "Find:∑k=1∞1k(k+1)(k+2)(k+3)"

Akbar Witt

Answered question

2022-02-22

Find:

\sum_{k = 1}^{\infty} \frac{1}{k (k + 1) (k + 2) (k + 3)}

meizhen85ulg

Beginner2022-02-23Added 6 answers

There is a general solution for this type of problem:

\sum_{k = 1}^{\infty} \frac{1}{k (k + 1) \cdot \dots \cdot (k + N)}

Observe that

\sum_{k \geq 1} \frac{1}{k (k + 1) \cdot \dots \cdot (k + N)} = \sum_{k \geq 1} \frac{(k - 1)!}{(k + N)!}

= \frac{1}{N!} \sum_{k \geq 1} \frac{Γ (k) Γ (N + 1)}{Γ (k + N + 1)} = \frac{1}{N!} \sum_{k \geq 1} B (k, N + 1)

Consequently,

\sum_{k = 1}^{\infty} \frac{1}{k (k + 1) \cdot \dots \cdot (k + N)} = \frac{1}{N!} \sum_{k \geq 1} \int_{0}^{1} x^{k - 1} {(1 - x)}^{N} d x

= \frac{1}{N!} \int_{0}^{1} (\sum_{k \geq 1} x^{k - 1}) {(1 - x)}^{N} d x

= \frac{1}{N!} \int_{0}^{1} {(1 - x)}^{N - 1} d x

= \frac{1}{N \cdot N!}

Here we have

N = 3

. Hence, the answer is

\frac{1}{3 \cdot 3!} = \frac{1}{18}

an2gi2m9gg

Beginner2022-02-24Added 9 answers

You can also go the hard way: write

\frac{1}{x (x + 1) (x + 2) (x + 3)} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x + 2} + \frac{D}{x + 3}

A (x + 1) (x + 2) (x + 3) + B x (x + 2) (x + 3) + C x (x + 1) (x + 3) + D x (x + 1) (x + 2) = 1

Now,
1. for

x = 0 : 6 A = 1

2. for

x = - 1 : - 2 B = 1

3. for

x = - 2 : 2 C = 1

4. for

x = - 3 : - 6 D = 1

Thus

\frac{1}{k (k + 1) (k + 2) (k + 3)} = \frac{1}{6} (\frac{1}{k} - \frac{3}{k + 1} + \frac{3}{k + 2} - \frac{1}{k + 3})

Hence your summation is

\frac{1}{6}

\sum_{k = 1}^{\infty} \frac{1}{k} - 3 \sum_{k = 1}^{\infty} \frac{1}{k + 1} + 3 \sum_{k = 1}^{\infty} \frac{1}{k + 2} - \sum_{k = 1}^{\infty} \frac{1}{k + 3} = \sum_{k = 1}^{\infty} \frac{1}{k} - 3 \sum_{k = 2}^{\infty} \frac{1}{k} + 3 \sum_{k = 3}^{\infty} \frac{1}{k} - \sum_{k = 4}^{\infty} \frac{1}{k}

= (1 + \frac{1}{2} + \frac{1}{3}) - 3 (\frac{1}{2} + \frac{1}{3}) + 3 \frac{1}{3}

= \frac{1}{3}

Therefore the sum of your series is

\frac{1}{6} \cdot \frac{1}{3} = \frac{1}{18}

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