Question

Find the limit of the following sequences or state that they diverge. $$\left\\{\frac{2 \tan ^{-1} n}{n^{3}+4}\right\\}$$

Short answer

Answer: The limit of the sequence as n approaches infinity is 0.

Step-by-step solution

Identify the function of the sequence and its limit

The given sequence is defined as follows: $$a_n = \frac{2 \tan^{-1} n}{n^3 + 4}$$ We are to find the limit of this sequence as n approaches infinity: $$\lim_{n \to \infty} a_n$$

Use properties of tangent function

Recall that \(\tan^{-1} n\) is the arc-tangent of n and increases monotonically with n as it lies in the interval \((-\frac{\pi}{2},\frac{\pi}{2})\). Therefore, we have: $$\lim_{n \to \infty} \tan^{-1} n = \frac{\pi}{2}$$

Compare the sequence with a simpler one

Since \(\tan^{-1} n\) converges to \(\frac{\pi}{2}\), we can compare the given sequence with a simpler one, such as: $$b_n = \frac{2 \cdot \frac{\pi}{2}}{n^3 + 4}$$ Which simplifies to: $$b_n = \frac{\pi}{n^3 + 4}$$ Now, we will find the limit of this new sequence.

Find the limit of the simpler sequence

Find the limit of \(b_n\) as n approaches infinity: $$\lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{\pi}{n^3 + 4}$$ As \(n\) approaches infinity, \(n^3 + 4\) increases without bound and the overall fraction goes to zero. Thus, the limit of this simpler sequence is: $$\lim_{n \to \infty} b_n = 0$$

Establish and apply the Sandwich Theorem

Notice that for all \(n\), $$0 \leq a_n \leq b_n$$ Since the limit of both \(a_n\) and \(b_n\) as n approaches infinity is 0, we can apply the Sandwich Theorem to conclude that the limit of the given sequence is also 0: $$\lim_{n \to \infty} a_n = 0$$