# calculus 2

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.

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1. 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.

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