Determine whether the given series are convergent. ∑ n = 1 ∞ n e n 2 \sum_{n=1}^{\infty}\frac{n^{e}}{n^2}

Now we'll move on to part c, where we are asked to determine whether the given series are convergent. Now looking at part c, here we have the sum from n equals one to infinity of n to the power of e over n squared. How can we figure out whether this is convergent or divergent? Well, right off the bat, this might not look like it's something super familiar to us that we're able to solve. But what I can do is play around with this to see if there's some sort of way that I can make this look familiar. And what I noticed is that we have the same base. We have n in both these cases. So n to the e over n squared can be rewritten. So this can also be written as n to the e times n to the negative two. I'm just taking that positive exponent in the denominator and making it a negative exponent to bring this n up to the numerator so we can multiply it by the e here. But what I can do is add these exponents. That's one of the rules that we've learned about. So because of this, we could say this is the same thing as n times e minus two. Now, this still doesn't look super familiar to us, but what I can actually try to do is force a p series here. And the way that I can do this is by making this exponent negative and bringing it to the denominator. So because of that, I can rewrite this n to the e minus two here as one over n to the negative e minus two. And negative e minus two, well, that's just going to be the same thing as two minus e. Now at this point, notice we do have a p series that emerges here. We have this p series because we have this two minus e, that's our p. For a p series, we're going to have it in this form where it's the sum from one over n to the p. In this case, I can see that my p is going to be this exponent, two minus e. Now in this case, two, we have this two, and then e is approximately 2.71. Now just looking at this, I can tell that's gonna leave us with an approximate value of negative zero. I'll write this as approximately, that these are both approximately here, for negative 0.71. And because of that, it's clear to me that p is going to be less than one. P has to be greater than one in order for it to converge, but since p is less than one, it diverges. So p has to be greater than one for it to converge, or if p is less than or equal to one, it diverges. And because of that, we could say that this series diverges. And that would be our solution right there to our example c. So this is how we can figure out whether a series converges or diverges. Sometimes we have to do a little bit of extra work to see the p series clearly. So I hope you found this helpful, and let's move on.

14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

Struggling with Calculus?

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Multiple Choice

Determine whether the given series are convergent.
$\sum_{n = 1}^{\infty} \frac{n^{e}}{n^{2}}$

Diverges because $p is less than 1$

Converges because $p is less than 1$

Diverges because $p is greater than 1$

Converges because $p is greater than 1$

Verified step by step guidance

Step 1: Recognize that the given series is $ \sum_{n=1}^{\infty} \frac{n^e}{n^2} $. This is a series where the general term is $ \frac{n^e}{n^2} $. To determine convergence, we need to analyze the behavior of the terms as $ n \to \infty $.

Step 2: Simplify the general term $ \frac{n^e}{n^2} $. Using the laws of exponents, rewrite it as $ n^{e-2} $. This allows us to focus on the exponent $ e-2 $ to determine the growth rate of the terms.

Step 3: Recall the p-series test, which states that a series of the form $ \sum_{n=1}^{\infty} \frac{1}{n^p} $ converges if $ p > 1 $ and diverges if $ p \leq 1 $. Here, the exponent $ e-2 $ plays the role of $ p $.

Step 4: Compare $ e-2 $ to 1. If $ e-2 > 1 $, the series converges. If $ e-2 \leq 1 $, the series diverges. Since $ e $ is a constant (approximately 2.718), calculate $ e-2 $ to determine whether it is greater than or less than 1.

Step 5: Conclude based on the value of $ e-2 $. If $ e-2 $ is less than or equal to 1, the series diverges because the terms do not decrease rapidly enough. If $ e-2 $ is greater than 1, the series converges because the terms decrease sufficiently fast.