Maths Q (1 Viewer)

[middlemarch]

Hi
i was hoping somebody could help me with this question,

(A) Use the area under the curve: y= 1/(1+u) for u=0 to u=X, to show that
X/(X+1) < ln(1+X) < X
this first part i can do ok

(B) Then use this inequality to show that:
(i) lim as (n-->infinity) of {[1 + (1/n)]^n } = e
(ii) e = 1 + 1/1! + 1/2! + 1/3! + ...

Thx

 

[gman03]

ln {[1 + (1/n)]^n }
= n * ln {[1 + (1/n)] }

let X = 1/n,
apply inequality

X/(X+1) < ln(1+X) < X

=> n * X/(X+1) < n * ln(1+X) < n * X
=> 1 / (1/n + 1) < ln {[1 + (1/n)]^n } < 1
=> e ^ { 1 / (1/n + 1) } < [1 + (1/n)]^n < e

n -> inf, then

=> e < lim (n->inf) {[1 + (1/n)]^n } < e

by pinching theorem,

lim (n->inf) {[1 + (1/n)]^n } = e

 

[gman03]

e = lim (n->inf) {[1 + (1/n)]^n }
= lim (n->inf) { 1 + nC1 / n + nC2 / n^2 + nC3 / n^3 + ...}

ought to prove

lim (n->inf) {nCr / n^r}
= lim (n->inf) { n! / [ r! ( n-r)! n^r ]}
= 1/r! * lim (n->inf) { n! / [( n-r)! n^r ]}
...
= 1/r!

so

e = lim (n->inf) { 1 + nC1 / n + nC2 / n^2 + nC3 / n^3 + ...}
= 1 + 1/1! + 1/2! + 1/3! ++ .....


Something like that

 

[KFunk]

(B) Then use this inequality to show that:
(i) lim as (n-->infinity) of {[1 + (1/n)]^n } = e
(ii) e = 1 + 1/1! + 1/2! + 1/3! + ...

Thx

If you expand using the binomial theorem then:

e = 1 + ⁿC₁(1/n) + ⁿC₂(1/n²) + ...

= 1 + n/(1!n) + [n(n-1)]/2!n² + [n(n-1)(n-3)]/3!n³ + ... + [nⁿ .... ]/n!nⁿ as n --> &infin;

If you take a limit of somthing like (x³ + 2x + 1)/(x³ + 3) as x approaches infinity you simply deal in the highest powers. Looking at the expansion you can see that you have n^Q ontop as your highest power and n^Q on the bottom as your highest power giving you 1/1 =1 . That's the best reasoning I could come up with.

hence e = 1 + 1/1! + 1/2! + 1/3! + ...

 

[middlemarch]

Thanks for the help gman and kfunk.
I have another question: what question number would you expect this to appear as?
M

 

[haboozin]

Thanks for the help gman and kfunk.
I have another question: what question number would you expect this to appear as?
M

i've seen this one,,,

7 or 8?

maybe 6 if they want to be slack

 

[middlemarch]

haboozin, thx.
Just making sure I'm not too out of it...