Expert: Steve Holleran - 4/9/2008

Question
QUESTION: Hi Steve,
I am working on sequences and would like help to mathematically find the limit of the sequence
sqrt(2), sqrt(2(sqrt(2)), sqrt(2sqrt(2(sqrt(2)))...
Thanks!

ANSWER: Hi Vince,

I'm not real good at sequences, but this is what I see:

a1= sqrt 2 = 2^(1/2)

a2 = sqrt(2*sqrt 2) = 2^(1/2) * (sqrt 2)^(1/2)

         = 2^(1/2) * [2^(1/2)]^(1/2)

         = 2^(1/2) * 2^(1/4)  = 2^(3/4)

a3 = 2^(7/8)

so continuing, I got that

an = 2^(2^n - 1) / 2^n)

now the exponent, (2^n-1) / 2^n = 2^n / 2^n - 1 / 2^n


         = 1 - [1 / 2^n]

And as n --> inf, 1/2^n -->0, so the sequence approaches

         2^1 = 2.


I hope this is okay.

Steve

---------- FOLLOW-UP ----------

QUESTION: How do I find if the sequence with nth term a(n) =(3^n)/n! converges?


Answer
Hi Vince,

Hey, you're really keeping me sharp on sequences!  I haven't worked with some of this stuff for awhile.

In any case, I think here you want to use the Ratio Test, which says that you want to form the ratio

         lim, n->inf   [ a(n+1) / a(n)]  

and if this is < 1, the sequence will converge.


So, here we have

   lim, n-> inf [3^(n+1)/(n+1)!] / [3^n / n!]

         = 3^(n+1)/ (n+1)! * n!/3^n

         = 3 * 3^n /(n+1)* n!  *  n! / 3^n

         = 3 / (n+1)

then  lim, n-> inf [3 / (n+1)] = 0 , so the sequence converges.


Steve