Thread: limits of sequences

find the limit of each sequence:
A) Sn = (1 + 1/2n)^(2n)
B) Sn = ( 1 + 1/n)^(2n)
C) Sn - (1 + 1/n) ^ (n-1)
D) Sn = (n/(n+1))^n
E) Sn = (1+ 1/2n)^n
F) Sn = ((n+2)/(n+1))^(n+3)

I found the answers to be this: A)e B) e^2 C) e D) 1/e E) sqrt(e) F) e...but I found the answers by using a different way then what the problems asks...it says to use the binomial theorem and the limit is referred to as e and is used as the base for natural logarithms.........Anyone know what they are asking and how to show the work needed to get the answers I got on A-F?

Originally Posted by learn18

find the limit of each sequence:
A) Sn = (1 + 1/2n)^(2n)
B) Sn = ( 1 + 1/n)^(2n)
C) Sn - (1 + 1/n) ^ (n-1)
D) Sn = (n/(n+1))^n
E) Sn = (1+ 1/2n)^n
F) Sn = ((n+2)/(n+1))^(n+3)

I found the answers to be this: A)e B) e^2 C) e D) 1/e E) sqrt(e) F) e...but I found the answers by using a different way then what the problems asks...it says to use the binomial theorem and the limit is referred to as e and is used as the base for natural logarithms.........Anyone know what they are asking and how to show the work needed to get the answers I got on A-F?

This is a little bit informal but the explanation is clearer this way.

A) By the binomial theorem:

Sn = (1 + 1/2n)^(2n) = 1 + (2n) (1/(2n) + (2n)(2n-1)(1/2n)^2/2! + ..

...................................... (2n)!/[(2n-r)! r!] (1/(2n)^r + ..

where r<=n.

Now as n becomes large the term (2n)!/[(2n-r)! is the product of r factors
of the form (2n-a) a = 0, 1, 2, .. r-1, which when divided by (2n)^r
approaches 1. So the r-th term of Sn approaches 1/r! as n goes to infinity.

Hence:

Lim(n->infty) Sn = 1 + 1/1! + 1/2! + ... + 1/r! + ... = e.

RonL

Here are B and C they are baby stuff.

..

Here is a last one and another attempt at #A.