Réponse :
Explications étape par étape :
Lemme:
[tex]\dfrac{1}{n}-\dfrac{1}{n+1} =\dfrac{n+1-n}{n(n+1)} =\dfrac{1}{n*(n+1)} \\\\\\\dfrac{1}{1*2} =\dfrac{1}{1} -\dfrac{1}{2} \\\\\dfrac{1}{2*3} =\dfrac{1}{2} -\dfrac{1}{3} \\\\\dfrac{1}{3*4} =\dfrac{1}{3} -\dfrac{1}{4} \\\\\dfrac{1}{4*5} =\dfrac{1}{4} -\dfrac{1}{5} \\\\...\\\dfrac{1}{(n-1)*n} =\dfrac{1}{n-1} -\dfrac{1}{n} \\\\\dfrac{1}{n*(n+1)} =\dfrac{1}{n} -\dfrac{1}{n+1} \\[/tex]
[tex]\dfrac{1}{1*2} +\dfrac{1}{2*3}+...+\dfrac{1}{n(n+1)} \\=\dfrac{1}{1} -\dfrac{1}{n+1} \\\boxed{=1 -\dfrac{1}{n+1} }\\=\dfrac{n+1-1}{n+1} =\dfrac{n}{n+1}[/tex]
