Determine whether the given series is convergent.∑n=1∞15n3\sum_{n=1}^{\infty}\frac{1}{^{^5}\sqrt{n^3}}

Since p ≤ 1 p\le1 , the series is divergent

Determine whether the given series is convergent. ∑ n = 1 ∞ 1 5 n 3 \sum_{n=1}^{\infty}\frac{1}{^{^5}\sqrt{n^3}}

We're asked to determine whether the given series are convergent, and we're starting here with example a. Now example a here has the sum from n equals one to infinity of one over the fifth root of n cubed. Now what I'm going to do is rewrite this sum, because I know that this sum from n equals one to infinity could also be rewritten as one over n cubed raised to the one fifth power, just changing that root right here. Now if I go ahead and continue to expand this out here, but continue to go over this, I should say, going step by step, I can multiply this n cubed, this three here with the one fifth. That's going to give me one over n to the three fifths power. Now this is where I'm gonna go ahead and stop. Because you may recognize at this point, we have a p series. And the p series takes this case where we have the sum from one over n to the p. So in this case, we have our p right here, which is this exponent, three fifths. Now we have a rule when it comes to this exponent. If p is greater than one, then we're going to have a case where the series is convergent. However, if p is less than or equal to one, then the series is divergent. Now, because I can see that we have three fifths, this means that we have this numerator that is less than the denominator. This is going to be a case where we are less than or equal to one. In this case, we're less than one, so our series diverges. So this right here will be the solution to example a. Hope you found this helpful, and let's move on to part b.

Determine whether the given series is convergent. ∑ n = 1 ∞ 1 5 n 3 \sum_{n=1}^{\infty}\frac{1}{^{^5}\sqrt{n^3}}

Since p ≤ 1 p\le1 , the series is divergent

Since p ≤ 1 p\le1 p ≤ 1 , the series is convergent

Since p > 1 p>1 , the series is convergent

Since p > 1 p>1 p > 1 , the series is divergent

14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Determine whether the given series is convergent.
$\sum_{n = 1}^{\infty} \frac{1}{^{^{5}} \sqrt{n^{3}}}$

Since $p\le1$ , the series is convergent

Since $p is greater than 1$ , the series is convergent

Since $p>1$ , the series is divergent

Since $p is less than or equal to 1$ , the series is divergent

Verified step by step guidance

Step 1: Recognize that the given series is ∑_{n=1}^{∞} 1 / (n^(3/5)). This is a p-series, which has the general form ∑_{n=1}^{∞} 1 / (n^p). The convergence or divergence of a p-series depends on the value of p.

Step 2: Recall the convergence criterion for p-series: If p > 1, the series converges. If p ≤ 1, the series diverges.

Step 3: Compare the exponent in the denominator of the given series (3/5) with the convergence criterion. Here, p = 3/5, which is less than 1.

Step 4: Based on the criterion, since p ≤ 1, the series diverges. This is because the terms of the series do not decrease quickly enough to sum to a finite value.

Step 5: Conclude that the series ∑_{n=1}^{∞} 1 / (n^(3/5)) is divergent because the exponent p = 3/5 satisfies the condition p ≤ 1.

5:44m

Watch next

Master Divergence Test (nth Term Test) with a bite sized video explanation from Patrick

Start learning

Convergence Tests practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice