# How do you test for convergence #(sin(2n))/(1+(2^n))# from n=1 to infinity?

Converges by the Direct Comparison Test.

We can use the Direct Comparison Test for this.

Now, we know the series

Then, since the larger series converges, so must the smaller series.

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

To test for convergence of the series (\frac{\sin(2n)}{1+2^n}) from (n=1) to (\infty), you can use the Limit Comparison Test. Let (a_n = \frac{\sin(2n)}{1+2^n}). Then, consider (b_n = \frac{1}{2^n}).

Take the limit as (n) approaches infinity of the ratio ( \frac{a_n}{b_n} ). If the limit is a finite positive number, then both series either converge or diverge together.

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

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

Now, since (-1 \leq \sin(2n) \leq 1) for all (n), we can say:

[ \lim_{n \to \infty} \frac{-2^n}{1+2^n} \leq \lim_{n \to \infty} \frac{\sin(2n)2^n}{1+2^n} \leq \lim_{n \to \infty} \frac{2^n}{1+2^n} ]

Both (\lim_{n \to \infty} \frac{-2^n}{1+2^n}) and (\lim_{n \to \infty} \frac{2^n}{1+2^n}) converge to (0). Therefore, by the Squeeze Theorem, (\lim_{n \to \infty} \frac{\sin(2n)2^n}{1+2^n} = 0).

Since the limit of the ratio is a finite positive number, by the Limit Comparison Test, both series either converge or diverge together. Thus, the series (\frac{\sin(2n)}{1+2^n}) converges.

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

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.

- How do you know if #sumn/(e(n^2))# converges from 1 to infinity?
- How do you determine the convergence or divergence of #Sigma ((-1)^nsqrtn)/root3n# from #[1,oo)#?
- How do you test for convergence given #Sigma (-1)^n(1-1/n^2)# from #n=[1,oo)#?
- How do you find #lim_(xto2) (x^3-6x-2)/(x^3-4x)# using l'Hospital's Rule or otherwise?
- How do you determine if the sum of #5^n/(3^n + 4^n)# from n=0 to infinity converges?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7