Expression de la somme \(\sum_{k=1}^{n} (\prod_{q=0}^{p} (\frac{1}{k+q})) \)

Cas particulier où \(p=1\) :

On cherche l'expression de la somme :
\(\sum_{k=1}^{n} \frac{1}{k(k+1)} = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + ... + \frac{1}{n(n+1)}\)
\(\forall n \in \mathbb{N}*, u_n = \frac{1}{n(n+1)}\)
\(u_n = \frac{(n+1)-n}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\)

\(S_n = u_1 + u_2 + ... + u_n\)
\(S_n = (\frac{1}{1} - \frac{1}{1+1}) + (\frac{1}{2} - \frac{1}{2+1}) + ... + (\frac{1}{n} - \frac{1}{n+1})\)
\(S_n = 1 - \frac{1}{n+1} = \frac{n}{n+1}\)

On a ainsi l'expression de la somme :
\(\sum_{k=1}^{n} \frac{1}{k(k+1)} = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + ... + \frac{1}{n(n+1)} = \frac{n}{n+1}\)

Cas particulier où \(p=2\) :

On cherche l'expression de la somme :
\(\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)} = \frac{1}{1 \times 2 \times 3} + \frac{1}{2 \times 3 \times 4} + ... + \frac{1}{n(n+1)(n+2)}\)
\(\forall n \in \mathbb{N}*, w_n = \frac{1}{n(n+1)(n+2)} = \frac{(n+2)-n}{n(n+1)(n+2)} \times \frac{1}{2}\)
\(w_n = \frac{1}{2} [\frac{1}{n(n+2)} - \frac{1}{(n+1)(n+2)}]\)

\(S_n = w_1 + w_2 + ... + w_n\)

On utilise la méthode de la somme télescopique :
\(S_n = \frac{1}{2} [\frac{1}{1(1+1)} - \frac{1}{(1+1)(1+2)}] + \frac{1}{2} [\frac{1}{2(2+1)} - \frac{1}{(2+1)(2+2)}] + ... + \frac{1}{2} [\frac{1}{(n-1)n} - \frac{1}{n(n+1)}] + \frac{1}{2} [\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}]\)
\(S_n = \frac{1}{2} [\frac{1}{2} - \frac{1}{(n+1)(n+2)}]\)
\(S_n = \frac{1}{4} - \frac{1}{2(n+1)(n+2)} = \frac{(n+1)(n+2)-2}{4(n+1)(n+2)}\)
\(S_n = \frac{n(n+3)}{4(n+1)(n+2)}\)

On a ainsi :
\(\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)} = \frac{1}{1 \times 2 \times 3} + \frac{1}{2 \times 3 \times 4} + ... + \frac{1}{n(n+1)(n+2)}\)
\(\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)} = \frac{n(n+3)}{4(n+1)(n+2)}\)

Démonstration dans le cas général :

\(\forall n \in \mathbb{N}*, \forall p \in \mathbb{N}*,\)
\(v_n = \frac{1}{n \times (n+1) \times ... \times (n+p)}\)
\(v_n = \frac{(n+p)-n}{n(n+1) \times ... \times (n+p)} \times \frac{1}{p}\)
\(v_n = \frac{1}{p} \times [\frac{1}{n(n+1) \times ... \times (n+p-1)} - \frac{1}{(n+1) \times (n+2) \times ... \times (n+p)}]\)

\(S_n = v_1 + v_2 + ... + v_n\)

On utilise une somme télescopique :
\(S_n = \frac{1}{p} [\frac{1}{1 \times (1+1) \times ... \times q} - \frac{1}{(1+1)(1+2) \times ... \times (1+p)}] + \frac{1}{p} [\frac{1}{2 \times (2+1) \times ... \times (p+1)} - \frac{1}{(2+1)(2+2) \times ... \times (2+p)}] + ... + \frac{1}{p} [\frac{1}{n \times (n+1) \times ... \times (n+p-1)} - \frac{1}{(n+1)(n+2) \times ... \times (n+p)}]\)
\(S_n = \frac{1}{p} [\frac{1}{1 \times 2 \times ... \times p} - \frac{1}{(n+1)(n+2) \times ... \times (n+p)}]\)
\(S_n = \frac{1}{p} [\frac{1}{p!} - \frac{n!}{(n+p)!}]\)
\(S_n = \frac{1}{p} \times \frac{(n+p)! - p! \times n!}{p! \times (n+p)!}\)

On a ainsi démontré que :
\(\sum_{k=1}^{n} (\prod_{q=0}^{p} (\frac{1}{k+q})) = \frac{1}{p} \times \frac{(n+p)! - p! \times n!}{p! \times (n+p)!}\)

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