# How do you do the limit comparison test for this problem #sqrt ( (n+1)/ (n^2+2))# as n goes to infinity?

Diverges when compared to

Then, both series diverge.

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The series diverges.

See work below:

The idea of the limit comparison test is that you essentially compare your unknown function to a function whose convergence you know (through another method: typically p-test). Here's how you do this:

Now, we just evaluate this limit using the same steps we learned in Calc 1. We just divide every term by the highest power:

Hope that helped :)

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To apply the limit comparison test for the series ( \sqrt{\frac{n+1}{n^2+2}} ) as ( n ) approaches infinity, follow these steps:

- Choose a known series that behaves similarly to the given series.
- Take the limit of the ratio of the given series and the chosen series as ( n ) approaches infinity.
- Determine the behavior of the limit:
- If the limit is a positive finite number, the series converge or diverge together.
- If the limit is 0 or infinity, the series diverge together.

After completing these steps, you'll be able to determine the convergence or divergence of the given series using the limit 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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- How do you find the sum of #Sigma(-2/7)^n# from n #[0,oo)#?
- How do you test for convergence for #sum(5^k+k)/(k!+3)# from k=1 to infinity?
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