Expert: Paul Klarreich - 10/10/2010

Question
hello, the question is : find the range of values of x>0 for which the series
(it has a sigma notation with n=1 at the bottom and infinity at the top)
x^n/(2^nn^2) is convergent...i have no idea so any help would be much
appreciated!

Answer
Questioner:   jay
Category:  Calculus
Private:  no
 
Subject:  converging function
Question:  hello, the question is : find the range of values of x>0 for which the series (it has a sigma notation with n=1 at the bottom and infinity at the top) x^n/(2^nn^2) is convergent...i have no idea so any help would be much appreciated!
....................................
    x^n
SUM ------  
   2^n n!


Ratio test:

Find lim[n>inf] of  a[n+1]/a[n]

As I recall, if this limit is < 1, it converges, if > 1, diverges, and = 1, try something else:
a[n+1]     x^n+1          2^n n!
------- = ------------- ----------
a[n]      2^n+1 (n+1)!  x^n

Lots of things cancel, so that is:
 x
------
2(n+1)

Now for any fixed x, that --> 0, so the series converges for all x.