Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you determine if $\sum \frac{n^{3}}{(n^{4}) - 1}$ $sum n^3/((n^4)-1)$ from n=2 to $n = \infty$ $n=oo$ is convergent?

1 Answer

Jun 14, 2015

It is divergent.

Explanation:

Since $\frac{n^{3}}{n^{4} - 1} ~ \frac{n^{3}}{n^{4}} ~ \frac{1}{n}$ $n^3/(n^4-1)~n^3/n^4~1/n$ that is the harmonic series, that diverges.

( $~$ $~$ means asymptotic).

Related questions

How do you use the limit comparison test on the series $\infty \sum n = 1 \frac{n}{2 n^{3} + 1}$ $sum_(n=1)^oon/(2n^3+1)$ ?
How do you use the limit comparison test on the series $\infty \sum n = 1 \frac{n + 1}{n \cdot \sqrt{n}}$ $sum_(n=1)^oo(n+1)/(n*sqrt(n))$ ?
How do you use the limit comparison test on the series $\infty \sum n = 2 \frac{\sqrt{n}}{n - 1}$ $sum_(n=2)^oosqrt(n)/(n-1)$ ?
How do you use the limit comparison test on the series $\infty \sum n = 1 \frac{n^{2} - 5 n}{n^{3} + n + 1}$ $sum_(n=1)^oo(n^2-5n)/(n^3+n+1)$ ?
How do you use the limit comparison test on the series $\infty \sum n = 1 \frac{1}{\sqrt{n^{3} + 1}}$ $sum_(n=1)^oo1/sqrt(n^3+1)$ ?
What is the Limit Comparison Test?
How do I use the Limit Comparison Test on the series $\infty \sum n = 1 sin (\frac{1}{n})$ $sum_(n=1)^oosin(1/n)$ ?
How do I know when to use limit comparison test vs the direct comparison test?
How do you use the comparison test (or the limit comparison test) for $\frac{1 + sin (x)}{10^{x}}$ $(1+sin(x))/10^x$ ?
How do you determine whether $\frac{1}{n!}$ $1/(n!)$ convergence or divergence with direct comparison test?

See all questions in Limit Comparison Test for Convergence of an Infinite Series

Impact of this question

8058 views around the world

You can reuse this answer
Creative Commons License