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

Answer 1

By comparing to #sum_(n=1)^oo1/n^2#, we find that #sum_(n=1)^oo1/(nsqrt(n^2+1))# is convergent through the limit comparison test.

The limit comparison test states that if #L=lim_(nrarroo)a_n/b_n# where #L# is a positive, finite value, then #suma_n# and #sumb_n# either both converge or both diverge.
When working with the limit comparison test, it's often helpful to let the series we are working with be #suma_n#. Thus, let #a_n=1/(nsqrt(n^2+1))#.
The trickier thing to do is choose #b_n#. The best thing to do is make #b_n# mimic whatever #a_n# does at infinity.
As #1/(nsqrt(n^2+1))# gets larger and larger, the constant becomes more and more irrelevant and we see that #a_n~~1/(nsqrt(n^2))=1/(n(n))=1/n^2#.
So, let #b_n=1/n^2#.
Now going to the limit comparison test, let #L=lim_(nrarroo)a_n/b_n#.
Then #L=lim_(nrarroo)(1/(nsqrt(n^2+1)))/(1/n^2)=lim_(nrarroo)n^2/(nsqrt(n^2+1))#
#=lim_(nrarroo)n/sqrt(n^2+1)#
#=lim_(nrarroo)n/(nsqrt(1+1/n^2)#
#=lim_(nrarroo)1/sqrt(1+1/n^2#
#=1/sqrt(1+0)#
#=1#
So #L=1#. Since this is positive and finite, we know that #suma_n# and #sumb_n# have the same convergence or divergence.
We know that #sumb_n=sum_(n=1)^oo1/n^2# converges through the p-series test, so this means that #suma_n=sum_(n=1)^oo1/(nsqrt(n^2+1))# also converges through the limit comparison test.
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Answer 2

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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Answer from HIX Tutor

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.

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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