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

Answer 1

#sum tan(1/n)# is divergent.

Let #a_n= tan(1/n)# and #b_n=1/n#.

Let us check the hypotheses of Limit Comparison Theorem.

#a_n>0# and #b_n>0# for all natural number #n#.
#lim_(n to infty)a_n/b_n=lim_(n to infty)tan(1/n)/(1/n)#

By l'Hospital's Rule,

#=lim_(n to infty)(sec^2(1/n)cdot cancel(-1/n^2)) /cancel(-1/n^2)=sec^2(0)=1#

Since #sum b_n# is a harmonic series (divergent), by Limit Comparison Test, we can conclude that #sum tan(1/n)# is also divergent.

I hope that this was clear.

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 determine the convergence of the series ( \sum \tan\left(\frac{1}{n}\right) ) from ( n = 1 ) to ( n = \infty ) using the Limit Comparison Test:

  1. Choose a series ( \sum b_n ) that you know the convergence of.
  2. Compute the limit ( \lim_{{n \to \infty}} \frac{\tan\left(\frac{1}{n}\right)}{b_n} ).
  3. If the limit is a positive finite number, then both series ( \sum \tan\left(\frac{1}{n}\right) ) and ( \sum b_n ) either both converge or both diverge.

Let's choose ( b_n = \frac{1}{n} ), a series we know to be divergent (it's the harmonic series).

Now, we calculate the limit:

[ \lim_{{n \to \infty}} \frac{\tan\left(\frac{1}{n}\right)}{\frac{1}{n}} ]

Using the limit ( \lim_{{x \to 0}} \frac{\tan(x)}{x} = 1 ), which is a well-known limit, we simplify:

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

Since this limit is a positive finite number (1), by the Limit Comparison Test, ( \sum \tan\left(\frac{1}{n}\right) ) and ( \sum \frac{1}{n} ) either both converge or both diverge.

Since we know that ( \sum \frac{1}{n} ) diverges (harmonic series), we conclude that ( \sum \tan\left(\frac{1}{n}\right) ) also diverges.

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