# 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#?

The series diverges.

Let's begin with an inequality :

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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.

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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.

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