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∑(k=1)→n k/(k+1)! .. some help please with this sum..
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\[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}\] hmmm
1/2 5/6 23/24 23/24 + 4/120 .... might need the wolf
1/2 5/6 23/24 23/24 + 4/120 .... might need the wol
you could have done this without the wolf if you spotted the pattern in the partial sums listed by @amistre64:\[\begin{align} k=1 &\qquad \frac{1}{2} &= \frac{2!-1}{2}!\qquad &= 1-\frac{1}{2!}\\ k=2 &\qquad \frac{5}{6} &= \frac{3!-1}{3!}\qquad &= 1-\frac{1}{3!}\\ k=3 &\qquad \frac{23}{24} &= \frac{4!-1}{4!}\qquad &= 1-\frac{1}{4!}\\ k=4 &\qquad \frac{119}{120} &= \frac{5!-1}{5!}\qquad &= 1-\frac{1}{5!}\\ ...\\ k=n &\qquad ... &= \frac{(n+1)!-1}{(n+1)!}\qquad &=1-\frac{1}{(n+1)!}\\ \end{align}\]
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