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JH

# Determine whether the series converges or diverges.$\displaystyle\sum_{n = 1}^{\infty} \frac {n + 3^n}{n+ 2^n}$

## DIVERGES

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##### Top Calculus 2 / BC Educators ##### Catherine R.

Missouri State University ##### Samuel H.

University of Nottingham ##### Michael J.

Idaho State University Lectures

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### Video Transcript

let's determine whether the Siri's conversions or diverges. So here, let's call this a M. And then let's define being to be three and over to the end. And now let's look at the limit. It's and goes to infinity of a n over bien. That's the limit as n goes to infinity. So now we multiplied by through the end over three and limiters and goes to infinity. That's end to and six end and then and three and plus six. And and now let's divide numerator and denominator by six in And then we'LL have the limit and goes to infinity. That will be an over three to the end, plus one over and over to to the n plus one. And now we can use the fact that and over three, the end equals one Excuse me equal zero. So here you could use Low Patel's rule thing, like they both go to infinity. So the derivative of the numerator is just one with respect to end the denominators three and Ellen three. And then the denominator goes toe infinities so that she zero and similarly and over to the inn also goes to zero. So we just have zero plus one over zero plus one equals one. Now the answer to our question the original Siri's will be the same is the answer for the sum of Bien. So the sum of the B end from one to infinity. That's just the sum. Let's just rewrite this as three over to the end, using your laws of exponents. But the Siri's diversions It's geometric with our equals three over, too. And that's bigger than one. Our equals three over two bigger than one. That means divergence, therefore, by the limit comparison test. Our Siri's also beverages. Lt's an equals one to infinity that's n plus three and in the numerator and plus two and in the denominator diverges as well. So that's our reasoning. And let me just point out before we end the video that the reason we were able to use the Lim comparison test was because the limit of ain't over bien. I was the number that satisfies this inequality here. It's positive number, so it can be zero, and I asked to be less than infinity, and that's what allows us to use the Lim comparison test. And that's our final answer

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##### Top Calculus 2 / BC Educators ##### Catherine R.

Missouri State University ##### Samuel H.

University of Nottingham ##### Michael J.

Idaho State University Lectures

Join Bootcamp