Mathematics, 22.04.2020 01:24 nicollexo21
Proof: let LaTeX: P\left(n\right)=\sum_{k=1}^n\frac{1 }{k(k+1)}=1-\frac{1}{n+1}.P ( n ) = β k = 1 n 1 k ( k + 1 ) = 1 β 1 n + 1 . Base case: P(1) = 1/2. Inductive step: suppose P(n) has already been proven for some arbitrary n. The statement P(n+1) is LaTeX: P\left(n+1\right)=\sum_{k=1}^{n+1}\ frac{1}{k\left(k+1\right)}=1-\frac{ 1}{n+2}P ( n + 1 ) = β k = 1 n + 1 1 k ( k + 1 ) = 1 β 1 n + 2 This concludes the proof by induction.
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Proof: let LaTeX: P\left(n\right)=\sum_{k=1}^n\frac{1 }{k(k+1)}=1-\frac{1}{n+1}.P ( n ) = β k = 1 n...
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