How do you use the limit comparison test on the series #sum_(n=1)^oon/(2n^3+1)# ?
Let us look at some details.
#lim_{n to infty}a_n/b_n=lim_{n to infty}n/{2n^3+1}cdotn^2/1 =lim_{n to infty}n^3/{2n^3+1}#
By signing up, you agree to our Terms of Service and Privacy Policy
To use the limit comparison test on the series (\sum_{n=1}^\infty \frac{n}{2n^3+1}), you need to compare it with a known convergent or divergent series. Choose a series (b_n) such that:
- (b_n > 0) for all (n),
- (\lim_{n \to \infty} \frac{a_n}{b_n}) exists and is a finite positive number.
Then, apply the limit comparison test:
- If (\sum_{n=1}^\infty b_n) converges, then (\sum_{n=1}^\infty a_n) also converges.
- If (\sum_{n=1}^\infty b_n) diverges, then (\sum_{n=1}^\infty a_n) also diverges.
In this case, we can choose (b_n = \frac{1}{n^2}). Now, we evaluate the limit:
[\lim_{n \to \infty} \frac{\frac{n}{2n^3+1}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^3}{2n^3+1} = \frac{1}{2}]
Since the limit exists and is a finite positive number, and (\sum_{n=1}^\infty \frac{1}{n^2}) converges (by p-series test with (p = 2 > 1)), by the limit comparison test, the given series (\sum_{n=1}^\infty \frac{n}{2n^3+1}) also converges.
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
- How do you determine if the improper integral converges or diverges #int ln(x)dx# from 0 to 2?
- How do you test the series #Sigma n^-n# from n is #[1,oo)# for convergence?
- How do you use the integral test to determine the convergence or divergence of the series from n=1 to infinity for #(arctan n) / (n^2 + 1)#?
- #sum_(n=1)^oo sin(n)/(n!)# How would i find if it converges or diverges?
- How do you find #lim (sqrtx-1)/(x-1)# as #x->1^+# using l'Hospital's Rule or otherwise?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7