# How do you use the limit comparison test to determine if #Sigma 1/(nsqrt(n^2+1))# from #[1,oo)# is convergent or divergent?

By comparing to

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To use the limit comparison test for the series Σ(1/(n√(n^2 + 1))) from n=1 to infinity, we need to compare it with a known convergent or divergent series.

Let's choose the series Σ(1/n^1.5), which is the harmonic series with the exponent increased by 0.5. This series is known to be divergent.

Now, we'll calculate the limit of the ratio of the terms of the two series as n approaches infinity:

lim (n → ∞) (1/(n√(n^2 + 1))) / (1/n^1.5)

After simplifying, we get:

lim (n → ∞) √(n^2 + 1) / n^0.5

Since the limit is equal to 1, by the limit comparison test, the series Σ(1/(n√(n^2 + 1))) and Σ(1/n^1.5) either both converge or both diverge.

Since we chose the series Σ(1/n^1.5) to be divergent, the original series Σ(1/(n√(n^2 + 1))) is also divergent.

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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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