Induction Question (1 Viewer)

[haboozin]

i. use mathematical induction to show that ln(n!) > n for positive integers n>=6

ok easy since ln7> 1 (dont need to answer this question for me)

ii. hense show that 1/n! < 1/e^n when n>=6

ok easy just rearange i (dont need to answer this either)


iii. Please answer this:
hence prove:
1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/8! ... < 103/60 + 1/e^5(e - 1)

 

[KFunk]

i. use mathematical induction to show that ln(n!) > n for positive integers n>=6

ok easy since ln7> 1 (dont need to answer this question for me)

ii. hense show that 1/n! < 1/e^n when n>=6

ok easy just rearange i (dont need to answer this either)


iii. Please answer this:
hence prove:
1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/8! ... < 103/60 + 1/e^5(e - 1)

1/n! < 1/eⁿ when n &ge; 6 so,

1/6! + 1/7! + ... < 1/e⁶ + 1/e⁷ + ... where the right hand site is geometric.

as n --> &infin;
1/e⁶ + 1/e⁷ + ... 1/eⁿ = (1/e⁶)/(1 - 1/e)
= 1/e⁵(e-1)

hence

1/6! + 1/7! + ... < 1/e⁵(e-1)

(then add 1/1! + 1/2! + ... + 1/5! to both sides)

1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/8! ... < 103/60 + 1/e^5(e - 1)

(since 1/1! + 1/2! + ... + 1/5! = 103/60)

 

[ishq]

From part (ii), we have -
1/n! < 1/e^n. [n >=6]

Therefore, in part (iii)

LHS
= 1/1! +1/2! + 1/3! +.... + 1/8! ...
< 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/e^6 + 1/e^7 + 1/e^8 ... (Using part (ii) )
< (add the first five) + 1/e^6 [ 1 + 1/e +1/e^2 + 1/e^3...] (Which is a geometic series)
< 103/60 + 1/e^6 [1/(1-1/e)] (Using Sum to Infinity)
< 103/60 + 1/e^5(e-1)
RHS


EDIT: Too Late...