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

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

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

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
but i got no idea how to intgegrate a factorial at the moment
 one year ago

amistre64 Group TitleBest 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
 one year ago

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

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
so the answer is 2e.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
good job :)
 one year ago
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