Friday, November 5, 2021
For every \(n\in\mathbb{N}\), \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}
\[\sum_{k=1}^{n}\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}
Base Case -> (n=1)
\frac{1}{(1+1)!}=1-\frac{1}{(1+1)!}
\frac{1}{2}=1-\frac{1}{2}
.5=.5
Induction Step -> Assume that for every \(n\in\mathbb{N}\)
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}
Since \frac{1}{2!}+...+\frac{n}{(n+1)!}+\frac{n+1}{((n+1)+1)!}=1-\frac{1}{((n+1)+1)!}
Then \frac{1}{2!}+...+\frac{n}{(n+1)!}+\frac{n+1}{(n+2)!}=1-\frac{1}{(n+2)!}
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