How do you use the limit comparison test for #sum ((sqrt(n+1)) / (n^2 + 1))# as n goes to infinity?
Remembering that:
so that series is convergent.
By signing up, you agree to our Terms of Service and Privacy Policy
Compare it to that using the limit comparison test:
By signing up, you agree to our Terms of Service and Privacy Policy
To use the limit comparison test for the series ( \sum \frac{\sqrt{n+1}}{n^2 + 1} ) as ( n ) goes to infinity, you need to find a series ( \sum b_n ) that you can compare it to. Here's how you do it:

Identify the problematic part of the series. In this case, it's ( \sqrt{n+1} ) in the numerator.

Find a simpler series ( \sum b_n ) where ( b_n ) is positive for all ( n \geq N ) for some ( N ) and is easier to evaluate.

Take the limit of the ratio of the two series as ( n ) approaches infinity:
[ \lim_{{n \to \infty}} \frac{a_n}{b_n} ]

If the limit is a finite positive number, then either both series converge or both series diverge. If the limit is zero or infinity, the comparison is inconclusive.
For this series, you can compare it to the series ( \sum \frac{1}{n^{3/2}} ) because ( \frac{\sqrt{n+1}}{n^{3/2}} ) is a simpler expression and easier to evaluate.
Taking the limit of the ratio:
[ \lim_{{n \to \infty}} \frac{\sqrt{n+1}}{n^{3/2}} = \lim_{{n \to \infty}} \frac{\sqrt{n+1}}{n^{3/2}} \cdot \frac{\frac{1}{\sqrt{n}}}{\frac{1}{\sqrt{n}}} = \lim_{{n \to \infty}} \frac{\sqrt{n+1}}{n} \cdot \frac{1}{\sqrt{n}} = 0 ]
Since the limit is a finite positive number (zero), by the limit comparison test, both series ( \sum \frac{\sqrt{n+1}}{n^2 + 1} ) and ( \sum \frac{1}{n^{3/2}} ) either converge or diverge together.
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a onesided 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 onesided 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 onesided 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 onesided 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.
 How do you use the limit comparison test to determine if #Sigma 2/(3^n5)# from #[1,oo)# is convergent or divergent?
 Identify whether the infinite series converge absolutely conditionally or dont #sum_(n=1)^oo (1)^(n+1) (n (arctan(n+1)arctan (n))# (Apply Mean Value theorem to conclude)?
 How do you determine whether the sequence #a_n=nn^2/(n+1)# converges, if so how do you find the limit?
 How do you determine the convergence or divergence of #Sigma (1)^(n+1)(1*3*5***(2n1))/(1*4*7***(3n2))# from #[1,oo)#?
 How do you use the limit comparison test for #sum( 3n2)/(n^32n^2+11)# as n approaches infinity?
 98% accuracy study help
 Covers math, physics, chemistry, biology, and more
 Stepbystep, indepth guides
 Readily available 24/7