How do you use basic comparison test to determine whether the given series converges or diverges for #sum n/sqrt(n^2-1)# from n=2 to #n=oo#?

Question

How do you use basic comparison test to determine whether the given series converges or diverges  for #sum n/sqrt(n^2-1)# from n=2 to #n=oo#?

Samantha Adkison · Accepted Answer

The series diverges.

Let's begin with an inequality :

#n^2 - 1 <= n^2#

#sqrt(n^2-1) <= sqrt(n^2) = n#, with #n>=1#

#1/(sqrt(n^2-1)) >= 1/n #, with #n >1#

#n/sqrt(n^2-1) >= n/n = 1#

#sum_{n=2}^infty(n/sqrt(n^2-1)) >= sum_{n=2}^infty(1)#

And obviously, #sum_{n=2}^infty(1) = +oo#, so it diverges.

By comparison test, #sum_{n=2}^infty(n/sqrt(n^2-1))# diverges too.

Cameron Alexander · Answer

undefined To determine the convergence or divergence of the series ∑ n/√(n^2 - 1) from n = 2 to n = ∞ using the basic comparison test, we compare it to a known series whose convergence or divergence is already established.

Here, we observe that as n approaches infinity, n/√(n^2 - 1) behaves similarly to 1/√n.

Since the series ∑ 1/√n from n = 1 to ∞ is a p-series with p = 1/2, and 1/2 > 0, it is known to converge.

By the basic comparison test, if 0 ≤ a_n ≤ b_n for all n and ∑ b_n converges, then ∑ a_n also converges.

As n approaches infinity, n/√(n^2 - 1) is bounded below by 1/√n, and since ∑ 1/√n converges, ∑ n/√(n^2 - 1) also converges by the basic comparison test.