Determine whether the given series is convergent. ∑ n = 1 ∞ n − 1 + n − 3 \sum_{n=1}^{\infty}n^{-1}+n^{-3}

So for part d of our problem, we have an interesting case. We're still trying to determine whether the given series are convergent, but now we have the sum from n equals one to infinity of n to the negative one plus n to the negative three. How can we go about solving this? Well, what I'm going to do is write this separately as two different sums. So the sum from n equals one to infinity of n to the negative one, which can be also written as one over n, just by taking that negative exponent there and writing it in the denominator, n to the first power. Now Now what I'm going to do is add that to the sum from n equals one to infinity of this here, n to the negative three, which can also be written as one over n cubed. Now notice that we get these two sums, which can both be found whether or not they're convergent by using a p series. Now we know that the p series is in this form where we have one over n to the p power. When p is less than one, this means our series is going to converge. This is the rule. And then when p or actually, no. Excuse me. That should be when p is less than or equal to one, the series is going to diverge. Now we know when p is less than or equal to one, the series is going to diverge. However, when p is greater than one, we know the series converges. So this is what we need to watch out for when doing these p series. Now I can clearly see in this case, we have one over n to the first power. This means that p is equal to one. This is also the harmonic series that we've seen, which we know always diverges. So this case down here is going to be divergent. Now, looking right here, I can see that we have a case where our p is equal to three. It's clear that three is greater than one. So because of that, we could say that this part of the series is going to converge. Now this is a little confusing because we have part of the series that converges, another part that diverges. But we know just by the rules of this, whenever you have something that that diverges, you added something to convert that converges, it doesn't actually matter. The whole series is just ultimately going to diverge. This diverging piece here is going to be dominant in this case. So our series indeed diverges. So because of that, this would be our solution to example d. So that's how you can solve these problems where you have to use a p series or a harmonic series. It's just about recognizing this form and possibly manipulating your original series that you have to get this form to come out so you can recognize whether the series will diverge or converge. Hope you found this video helpful.

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14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

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Multiple Choice

Determine whether the given series is convergent.
$\sum_{n = 1}^{\infty} n^{- 1} + n^{- 3}$

Diverges

Converges

Both converges and diverges

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Verified step by step guidance

Step 1: Understand the problem. The series given is $ \sum_{n=1}^{\infty} n^{-1} + n^{-3} $. We need to determine whether this series converges or diverges. To do this, analyze each term separately: $ n^{-1} $ and $ n^{-3} $.

Step 2: Recall the p-series test. A p-series is of the form $ \sum_{n=1}^{\infty} \frac{1}{n^p} $, and it converges if $ p > 1 $ and diverges if $ p \leq 1 $. Apply this test to each term in the series.

Step 3: Analyze $ n^{-1} $. Here, $ p = 1 $. Since $ p \leq 1 $, the series $ \sum_{n=1}^{\infty} n^{-1} $ diverges. This is known as the harmonic series, which is a classic example of divergence.

Step 4: Analyze $ n^{-3} $. Here, $ p = 3 $. Since $ p > 1 $, the series $ \sum_{n=1}^{\infty} n^{-3} $ converges. This term alone would contribute to convergence.

Step 5: Combine the results. The series $ \sum_{n=1}^{\infty} n^{-1} + n^{-3} $ is the sum of a divergent series ($ n^{-1} $) and a convergent series ($ n^{-3} $). Since the divergence of $ n^{-1} $ dominates, the entire series $ \sum_{n=1}^{\infty} n^{-1} + n^{-3} $ diverges.