# How do you use the limit comparison test to determine if #sum_(n=3)^(oo) 3/sqrt(n^2-4)# is convergent or divergent?

See below.

We know that

Harmonic series. See https://tutor.hix.ai(mathematics)

Now considering

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To use the limit comparison test, select a series you believe to be similar to the given series. In this case, let's choose the series ( \sum_{n=1}^{\infty} \frac{1}{n} ), a well-known divergent series. Then, you'll calculate the limit of the quotient of the two series as ( n ) approaches infinity:

[ \lim_{n \to \infty} \frac{\frac{3}{\sqrt{n^2 - 4}}}{\frac{1}{n}} ]

Simplify this expression by multiplying both the numerator and denominator by ( \frac{1}{n} ) and then applying limit properties. After simplification, if the limit is a positive finite number, then both series behave similarly, either both convergent or both divergent. If it's zero or infinity, it means they behave differently.

After solving the limit, if the limit is a positive finite number, then both series behave similarly, either both convergent or both divergent. If it's zero or infinity, it means they behave differently. Depending on the outcome, you can conclude whether the original series ( \sum_{n=3}^{\infty} \frac{3}{\sqrt{n^2-4}} ) is convergent or divergent accordingly.

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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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- How do you determine whether the infinite sequence #a_n=n+1/n# converges or diverges?
- How do you find the interval of convergence of #Sigma (x+10)^n/(lnn)# from #n=[2,oo)#?

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