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amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0if there was a pattern we could find from the partial sums, that could be useful; but its still a shot in the dark

Rachel123
 2 years ago
Best ResponseYou've already chosen the best response.0Can i use the fact that the sum of 1/k! from 0 to infinity is e1

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0i cant say for sure .... you can give it a shot and then verify it with the wolf

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0you can try taking the partial sums up to some arbitrary point, and then averaging the integrals of the remainder

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0but i got no idea how to intgegrate a factorial at the moment

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0\[e=\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}+\frac1{5!}+...\] \[k^2e=k^2\left(\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}+\frac1{5!}+...\right)\] pfft, theres prolly something wrong with my idea on that one

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.2\[\large \frac{ k^2 }{ k! }=\frac{k(k1+1)}{k!}=\frac{k(k1)}{k!}+\frac{k}{k!}=\frac{1}{(k2)!}+\frac{1}{(k1)!}\]
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