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# Test the series for convergence or divergence.$\displaystyle \sum_{n = 1}^{\infty} \frac {1 \cdot 3 \cdot 5 \cdot \cdot \cdot \cdot \cdot (2n - 1)}{2 \cdot 5 \cdot 8 \cdot \cdot \cdot \cdot \cdot (2n - 1)}$

## convergent

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okay for this problem. I believe there's supposed to be a three here, not a two. So that might be one of the hardest problems for some people just recognizing this pattern down here. So we'll use the ratio test here. These terms here being r a in terms. So we're looking at limit as n goes to infinity of absolute value of and plus one over n. These terms of positive says the same thing as without the absolute value here. Okay, so this is limit as n goes to infinity one times, three times five times. Not that that time two and minus one And then we're looking at and plus one here. So we go one step further and similarly down below. We have two times, five times eight times that. That that And then we have times three and minus one and we're looking at a seven plus wan. So we go one step more than that. So all of this is just our ace of n plus one and we're dividing by a seven, which is the same thing as multiplying by the reciprocal of Ace have been so multiplying by two times five times a day, Not all the way up to three and minus one and then down below. We're having one times three times, five times that thought Times lips shed parentheses here, times two n minus one. Okay, so that one times three times, five times two in minus one. All of that is going to cancel out with all of this. Similarly, two times, five times, eight times. Not that thought Times three and minus one. All of that is going to cancel that with all of this. So now we have a limit as n goes to infinity. Two times in plus one minus one over three times in plus one minus one. And this limo is two thirds, which is less than one. So we do have convergence.

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