How do you test the series #Sigma (3n^2+1)/(2n^4-1)# from n is #[1,oo)# for convergence?

Question

Aurora Alvarado · Accepted Answer

undefined To test the convergence of the series $ \sum_{n=1}^\infty \frac{3n^2+1}{2n^4-1} $, you can use the ratio test.

Apply the ratio test:
1. Calculate the limit as $ n $ approaches infinity of the absolute value of the ratio of successive terms:
$$ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$
where $ a_n = \frac{3n^2+1}{2n^4-1} $.

2. Evaluate the limit:
$$ L = \lim_{n \to \infty} \left| \frac{\frac{3(n+1)^2+1}{2(n+1)^4-1}}{\frac{3n^2+1}{2n^4-1}} \right| $$

3. Simplify and compute the limit $ L $.

4. If $ L < 1 $, the series converges absolutely.
   If $ L > 1 $, the series diverges.
   If $ L = 1 $, the test is inconclusive.

This procedure will determine the convergence or divergence of the given series.

William Abdalla · Answer

#sum_(n=1)^oo(3n^2+1)/(2n^4-1)#

is convergent based on the direct comparison test.

We can test the series by direct comparison. As: #sum_(n=1)^oo a_n = sum_(n=1)^oo(3n^2+1)/(2n^4-1)# is a series with positive terms, we need to find a convergent series #sum_(n=1)^oo b_n# such that: #a_n < b_n# for #n>N# Now we have: #(3n^2+1)/(2n^4-1) < (3n^2)/(2n^4-1) < (3n^2)/(2n^4) < 1/n^2# But #sum_(n=1)^oo 1/n^2 # is convergent based on the p-series test, so also: #sum_(n=1)^oo(3n^2+1)/(2n^4-1)# is convergent