Question
Test the series for convergence/divergence.
The Summation from n=1 to infinity of:
1*3*5...(2n-1)/(2*5*8...(3n-1)
I'm not sure what to do with the extra terms on the left.
The Summation from n=1 to infinity of:
1*3*5...(2n-1)/(2*5*8...(3n-1)
I'm not sure what to do with the extra terms on the left.
Answers
The nth term in the summation can be written as:
1*3*5...(2n-1)/(2*5*8...(3n-1) =
Product i = 1 to n of (2 i-1)/(3i-1)
Then because
3/2 (2i-1) = 3i - 3/2 < 3i-1
we have that:
(2 i-1)/(3i-1) < 2/3
Therefore:
Product i = 1 to n of (2 i-1)/(3i-1) <
(2/3)^n
The summation of (2/3)^n from n = 1 to infinity converges, therefore the original summation converges.
1*3*5...(2n-1)/(2*5*8...(3n-1) =
Product i = 1 to n of (2 i-1)/(3i-1)
Then because
3/2 (2i-1) = 3i - 3/2 < 3i-1
we have that:
(2 i-1)/(3i-1) < 2/3
Therefore:
Product i = 1 to n of (2 i-1)/(3i-1) <
(2/3)^n
The summation of (2/3)^n from n = 1 to infinity converges, therefore the original summation converges.
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