Given that `n!>=2^(n-1)` Deduce thatÂ `sum_(k=1)^n 1/(k!) <= 2-1/2^(n-1)` Hence show that e<=3, where e if the base of natural logarithms.
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`sum_(k=1)^n 1/(k!) <= sum_(k=1)^n 1/2^(k-1)`
`n!>=2^(n-1)`
Since n is positive;
`1/(n!)<=1/2^(n-1)`
` ` Get the summation at both sides.
`sum_(k=1)^n 1/(k!) <= sum_(k=1)^n 1/2^(k-1)`
`sum_(k=1)^n 1/2^(k-1) = 1+(1/2)+(1/2)^2+......+(1/2)^(k-1)`
`sum_(k=1)^n (1/2)^(k-1)` represent a sum of a geometric series.
`sum_(k=1)^n 1/2^(k-1) = (1-(1/2)^n)/(1-1/2) = 2-2/2^n`
`sum_(k=1)^n 1/(k!) <= sum_(k=1)^n 1/2^(k-1)`
`sum_(k=1)^n 1/(k!) <= 2-2/2^n`
`sum_(k=1)^n 1/(k!) <= 2-1/2^(n-1)`
So the answer is proved.
We know that;
`e=lim_(n->oo) sum_(k=0)^n 1/(k!) `
`e= lim_(n->oo) [1+sum_(k=1)^n 1/(k!)]`
`e=1+ lim_(n->oo) sum_(k=1)^n 1/(k!)`
We proved that;
`sum_(k=1)^n (1/k!) <= 2-1/2^(n-1)`
`lim_(n->oo) sum_(k=0)^n (1/k!)<=1+ lim_(n->oo) [2-1/(2^(n-1))]`
`e<=1+2-0`
`3<=3`
So the answer is proved.
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