Evaluate `sum_(n=1)^(infty)1/(n!) `
The series converges from the ratio test:
`lim_(n->oo)(1/((n+1)!))/(1/(n!))=lim_(n->oo)1/(n+1)=0 `
The sequence of partial sums:
n=1 1
n=2 1+1/2=1.5
n=3 1+1/2+1/6=5/3 or `1.bar(6) `
n=4 1+1/2+1/6+1/24=`41/24=1.708bar(3) `
n=5 1+1/2+1/6+1/24+1/120=`103/60=1.71bar(6) `
This sequence is approaching e-1 where e is the base of the natural logorithm and `e-1~~1.718281828... `
Comments
Post a Comment