How do you use the limit comparison test to determine whether the following converge or diverge given #sin(1/(n^2))# from n = 1 to infinity?

Answer 1

#color(red)(sum_(n=1)^∞ sin(1/n^2)" is convergent")#.

#sum_(n=1)^∞ sin(1/n^2)#
The limit comparison test states that if #a_n# and #b_n# are series with positive terms and if #lim_(n→∞) (a_n)/(b_n)# is positive and finite, then either both series converge or both diverge.
Let #a_n = sin(1/n^2)#
Let's think about the end behaviour of #a_n#.
For large #n#, the argument #1/n^2# becomes small.

We can use the small-angle approximation:

As #x→0, sin x → x#.
So, for large #n#, #a_n# acts like #1/n^2#.
Let #b_n= 1/n^2#.
Then #lim_(n→∞)a_n/b_n = lim_(n→∞)sin(1/n^2)/(1/n^2)= lim_(n→∞)(1/n^2)/(1/n^2) = 0/0#

This indeterminate result is discouraging, but we can apply L'Hôpital's rule:

If #lim_(x→a)f(x)/g(x) =0/0 or ±∞/∞#, then #lim_(x→a)f(x)/g(x)= lim_(x→a) (f'(x))/(g'(x))#.
Let #x = 1/n^2#
So, #lim_(x→0)sinx/x = lim_(x→0)cosx/1 =1/1=1#
But #b_n= 1/n^2# is convergent, so
#a_n = sin(1/n^2)# is also convergent.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To use the limit comparison test for determining convergence or divergence of the series ( \sum \sin\left(\frac{1}{n^2}\right) ) from ( n = 1 ) to ( \infty ), you need to compare it with a known series that converges or diverges.

Let's denote the given series as ( a_n = \sin\left(\frac{1}{n^2}\right) ).

Choose a series ( b_n ) that you know the convergence or divergence of, and that behaves similarly to ( a_n ) as ( n ) approaches infinity. In this case, a good choice is ( b_n = \frac{1}{n^2} ), as it has a similar form to ( a_n ).

Next, compute the limit:

[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\sin\left(\frac{1}{n^2}\right)}{\frac{1}{n^2}} ]

If this limit is a positive finite number, then both series ( \sum a_n ) and ( \sum b_n ) converge or diverge together. If the limit is zero or infinity, the series may converge or diverge independently.

Calculate the limit and determine its value:

[ \lim_{n \to \infty} \frac{\sin\left(\frac{1}{n^2}\right)}{\frac{1}{n^2}} = \lim_{n \to \infty} n^2 \sin\left(\frac{1}{n^2}\right) = 1 ]

Since the limit is a positive finite number (1), we can conclude that ( \sum \sin\left(\frac{1}{n^2}\right) ) and ( \sum \frac{1}{n^2} ) either both converge or both diverge.

Since ( \sum \frac{1}{n^2} ) is a convergent series (it's a special case of the p-series with ( p = 2 )), by the limit comparison test, we can conclude that ( \sum \sin\left(\frac{1}{n^2}\right) ) also converges.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7