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Numerade Educator

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Problem 16 Easy Difficulty

Test the series for convergence or divergence.

$ \displaystyle \sum_{n = 1}^{\infty} \frac {\sqrt{n^4 + 1}}{n^3 + n} $

Answer

Diverges

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Video Transcript

in this problem, we have to to determine the convergence or divergence of the Siri's, and specifically we have to determine the test to use. We're not told the test to use. So let's first start out with what we're given. Were given the Siri's from n equals one to infinity off the square root of Enter the fourth plus one all over and cubed plus end well, In this case, we can use the limit comparison test, and we can compare it with beasts of n equals one over end. Now, why are we using one over an? Well, we know the convergence or divergence of the Siris of one over N that's known We can use that to compare the convergence or divergence diversions of the Siri's were given so we can take the limit is an approaches infinity of a seven over beasts of men. A Saban is what we're given, and Visa Ban is what we're comparing it to. So we'll have the limits and approaches infinity of the square root event to the fourth plus one all over and cubed plus n. Then we'll divide that whole quantity by one over end so we can do a little bit of simplification. Remember, dividing by a fraction is like multiplying by the reciprocal. So we'll get the limit as an approaches infinity of end times, the square root of end of the fourth plus one all over and cubed plus n. And now we could solve that like a normal limit. What we'll do is we'll divide the numerator and the denominator bye and cubed. When we do that, we'll get the limit. As an approaches Infinity off the square root of one plus one over entered the fourth, divided by one plus one over and squared. And now we're in a position where we can plug in infinity and we do that. We get the square root of one plus zero all over one plus zero, and that is equivalent toe one. Now this is important because we got 11 is not zero, and one is not infinity. This is a definitive number that tells us that convergence or divergence of our Siri's. So what we can determine is that because the Siri's from an equals one to infinity of one over end is a divergent Siri's, and we got one as the answer to our limit, the Siri's that were given bridge by the limit comparison test. So I hope that this problem helped you understand a little bit more about evaluating the convergence or divergence of a serious and how we can use the limit comparison tests to evaluate that.