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Match the power series with its interval of convergence:

1.Sigma(sum of) (infinity to n = 1)
n! (3x-5)^n/5^n

2. Sigma(sum of) (infinity to n = 1)
(x-5)^n/(n!)5^n

3. Sigma(sum of) (infinity to n = 1)
(3x)^n/n^5

4. Sigma(sum of) (infinity to n = 1)
(x-5)^n/(5)^n

(-1/3, 1/3)
(-infinity, infinity)
(0,10)
{5/3}

Thanks of any help with this problem. I am not necessarily looking of the answers, just how to set up the problems. Thank you!

2007-04-26 14:28:42 · 1 answers · asked by Mrs.Sizemore 2 in Science & Mathematics Mathematics

1 answers

Take absolute value and then apply the ratio test
1)a_n+1 /a_n= (n+1)I3x+5I/5 which has limit + infinity.So lim a_n is not zero and the series diverges for all x (except 3x+5=0)where all terms are 0
4)a_n =(Ix-5)/5I)^n is a geometric series with r =Ix-5)/5I so if
I(x-5)/5I<1 is convergent and divergent for all other values

-1<(x-5)/5 <1 so -5 and
0 I think you can do the others

2007-04-26 14:48:11 · answer #1 · answered by santmann2002 7 · 0 0

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