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Limits of sequences Find the limit of the following sequences or determine that the sequence diverges. $$\left\\{\ln \left(n^{3}+1\right)-\ln \left(3 n^{3}+10 n\right)\right\\}$$

Short Answer

Expert verified
Question: Determine if the following sequence converges and, if so, calculate its limit: $$\lim_{n\to\infty} \ln \left(n^{3}+1\right)-\ln \left(3 n^{3}+10 n\right)$$ Answer: This sequence converges, and its limit is \(\ln\left(\frac{1}{3}\right)\).

Step by step solution

01

Rewrite the logarithm difference as a division

Using the properties of logarithms, the difference of two logarithms can be rewritten as the logarithm of a quotient. We have: $$\ln \left(n^{3}+1\right)-\ln \left(3 n^{3}+10 n\right) = \ln\left(\frac{n^{3}+1}{3 n^{3}+10 n}\right)$$
02

Simplify the expression inside the logarithm

To make it easier to find the limit, we can simplify the expression inside the logarithm. Divide both the numerator and the denominator by \(n^3\): $$\ln\left(\frac{n^{3}+1}{3 n^{3}+10 n}\right) = \ln\left(\frac{\dfrac{n^3}{n^3} + \dfrac{1}{n^3}}{\dfrac{3n^3}{n^3} + \dfrac{10n}{n^3}}\right) = \ln\left(\frac{1+\dfrac{1}{n^3}}{3+\dfrac{10}{n^2}}\right)$$
03

Calculate the limit of the sequence

Now, we will calculate the limit of the sequence as \(n\) goes to infinity: $$\lim_{n\to\infty} \ln\left(\frac{1+\dfrac{1}{n^3}}{3+\dfrac{10}{n^2}}\right)$$ As \(n\) goes to infinity, the terms \(\dfrac{1}{n^3}\) and \(\dfrac{10}{n^2}\) go to 0, so the limit will be: $$\lim_{n\to\infty} \ln\left(\frac{1}{3}\right)$$ Since the logarithm is continuous, we can directly apply the limit to the argument inside the logarithm: $$\ln\left(\lim_{n\to\infty} \frac{1}{3}\right) = \ln\left(\frac{1}{3}\right)$$ So, the limit of the sequence is: \(\boxed{\ln\left(\frac{1}{3}\right)}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Divergence of sequences
When dealing with sequences, it's important to understand whether a sequence converges or diverges. For a sequence to converge, it must approach a specific value as its terms go to infinity. On the other hand, a sequence diverges if it does not settle down to a fixed value.
In our example, we're checking if the sequence given by \(\ln \left(n^{3}+1\right)-\ln \left(3 n^{3}+10 n\right)\) diverges or not.Sequences can diverge in several ways:
  • They might head towards infinity or negative infinity.
  • They could oscillate without settling down.
  • The terms might become erratic without repeating or settling down.
Fortunately, with logarithms, the sequence tends to move predictably. By managing the inner expressions and applying limit calculations, you determine the behavior of the sequence. If it hones into a particular value, like in our task, it doesn't diverge but converges instead.
Logarithmic identities
Logarithmic identities are powerful tools for simplifying and manipulating expressions. They make tasks like limit calculations much simpler.Consider the identity involved in our exercise: the difference of two logarithms leads to the logarithm of a quotient:\[ \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \]This property allows us to transform complex expressions into simpler forms, paving the way for easier limit calculations.
In the given sequence, \(\ln \left(n^{3}+1\right)-\ln \left(3 n^{3}+10 n\right)\) became \(\ln\left(\frac{n^{3}+1}{3 n^{3}+10 n}\right)\), which is a simpler form, letting us directly engage in further simplification.Logarithmic identities help:
  • Simplify expressions.
  • Convert complex differences into calculable quotients.
  • Handle products as single expressions or sums.
Once simplified, the effects of limits are more apparent, making your calculations cleaner and more efficient.
Limit calculation
Limit calculations are essential in determining the behavior of functions and sequences at boundary points, often involving "as \(n\) approaches infinity."
In the discussed exercise, the limit focused on evaluating the behavior of \(\ln\left(\frac{1+\frac{1}{n^3}}{3+\frac{10}{n^2}}\right)\) as \(n\) becomes infinitely large.Steps for calculating this limit included:
  • Recognizing terms like \(\frac{1}{n^3}\) and \(\frac{10}{n^2}\) approach zero as \(n\) grows.
  • Understanding the limit simplifies into \(\ln\left(\frac{1}{3}\right)\) as other components diminish.
The concept of limit allows us to handle infinite processes, solving them in definitively simple terms. In this case,
it assured us that the sequence converged to \(\ln\left(\frac{1}{3}\right)\). Understanding limit calculations is crucial in calculus, serving as a bridge between algebra and the world of functions changing over time.

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Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer \(N\) and call it \(a_{0} .\) This is the seed of a sequence. The rest of the sequence is generated as follows: For \(n=0,1,2, \ldots\) $$a_{n+1}=\left\\{\begin{array}{ll} a_{n} / 2 & \text { if } a_{n} \text { is even } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is odd } \end{array}\right.$$ However, if \(a_{n}=1\) for any \(n,\) then the sequence terminates. a. Compute the sequence that results from the seeds \(N=2,3\) 4,...., 10. You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers \(N\), the sequence terminates after a finite number of terms. b. Now define the hailstone sequence \(\left\\{H_{k}\right\\}\), which is the number of terms needed for the sequence \(\left\\{a_{n}\right\\}\) to terminate starting with a seed of \(k .\) Verify that \(H_{2}=1, H_{3}=7,\) and \(H_{4}=2\) c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?

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