# 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.

### 1 Answer | Add Yours

`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.*

**Sources:**