Question

Is the sequence bounded, monotone, and convergent or divergent? If it is convergent, find the limit. $$ a_{n}=\frac{\ln (n+1)}{\sqrt{n+1}} $$

Short answer

The sequence is bounded, not monotone, and convergent with limit 0.

Step-by-step solution

Check if the sequence is bounded

To determine if the sequence \( a_n = \frac{\ln(n+1)}{\sqrt{n+1}} \) is bounded, consider the behavior of \( \ln(n+1) \) and \( \sqrt{n+1} \). \( \ln(n+1) \) increases without bound as \( n \to \infty \), but more slowly than \( \sqrt{n+1} \), which also increases. For small \( n \), evaluate a few terms: \( a_1 = \frac{\ln(2)}{\sqrt{2}} \approx 0.49 \), and \( a_2 = \frac{\ln(3)}{\sqrt{3}} \approx 0.63 \). Since both the numerator and denominator increase, and the denominator grows faster, the sequence is bounded for large \( n \).

Check if the sequence is monotone

Examine if the sequence \( a_n = \frac{\ln(n+1)}{\sqrt{n+1}} \) is increasing or decreasing. Consider the ratio of terms \( \frac{a_{n+1}}{a_n} = \frac{\ln(n+2)\sqrt{n+1}}{\ln(n+1)\sqrt{n+2}} \). This is complex to determine analytically if it is always less than 1; so, consider numerically checking values: \( a_1 < a_2 < a_3 \) proves true for early terms, suggesting non-monotonic behavior due to varying rates of growth of \( \ln(n+1) \) versus \( \sqrt{n+1} \). The sequence is not strictly monotone.

Check for convergence and find the limit

Using limits, evaluate \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\ln(n+1)}{\sqrt{n+1}} \). Use L'Hopital's Rule because both the numerator and denominator approach infinity: differentiate \( \ln(n+1) \) to get \( \frac{1}{n+1} \) and differentiate \( \sqrt{n+1} \) to get \( \frac{1}{2\sqrt{n+1}} \). Thus, \( \lim_{n \to \infty} \frac{1/(n+1)}{1/[2\sqrt{n+1}]} = \lim_{n \to \infty} \frac{2\sqrt{n+1}}{n+1} = \lim_{n \to \infty} \frac{2}{\sqrt{n+1}} = 0 \). Thus, the sequence converges to 0.

Key concepts

Bounded Sequence

A bounded sequence is one where all its terms are confined within a specific range.
In other words, there exist numbers, say, a lower bound and an upper bound such that every term in the sequence falls between these two numbers.
This doesn't mean the sequence stops growing, but simply that the growth is restricted within certain limits over the entire sequence. To see this in action, consider the sequence given in the problem statement: \[ a_{n} = \frac{\ln (n+1)}{\sqrt{n+1}} \]As the denominator \( \sqrt{n+1} \) increases faster than the numerator \( \ln(n+1) \), the overall value of \( a_n \) will not increase indefinitely.
Thus, even as \( n \) grows very large, \( a_n \) remains within a fixed range, making the sequence bounded from above by zero.
This kind of behavior is what qualifies it as a bounded sequence; while the numerator and denominator both head towards infinity, their rates control the limits of the sequence's values.

Monotone Sequence

A monotone sequence is defined by its consistent behavior in terms of either continuously increasing or decreasing.
In simpler terms, if a sequence keeps going up without dropping in between, it's increasing and vice versa.Examining the given sequence:\[ a_{n} = \frac{\ln (n+1)}{\sqrt{n+1}} \]Checking whether this sequence is monotone involves determining if each term is consistently greater or lesser than the previous one.
While small values such as \( a_1 \approx 0.49, a_2 \approx 0.63 \) show an increasing tendency, this does not ensure it's always increasing or decreasing, especially for large \( n \).Therefore, the sequence demonstrates non-monotonic behavior because the ratios \( \ln(n+1) \) versus \( \sqrt{n+1) \) do not consistently lead to one term being greater or smaller than the next.
This means the sequence does not maintain a strict direction of increase or decrease throughout.

L'Hopital's Rule

L'Hopital's Rule is a tool in calculus used to evaluate limits of indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
This rule transforms complicated limits into simpler forms, making them solvable.For the exercise in question, we apply L'Hopital's Rule as follows:The limit of our sequence as \( n \to \infty \):\[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\ln(n+1)}{\sqrt{n+1}} \]is initially \( \frac{\infty}{\infty} \).
This is where L'Hopital's Rule comes into play, allowing us to differentiate the numerator and denominator separately:- The derivative of \( \ln(n+1) \) is \( \frac{1}{n+1} \).- The derivative of \( \sqrt{n+1} \) becomes \( \frac{1}{2\sqrt{n+1}} \).Applying L'Hopital’s Rule simplifies the limit to:\[ \lim_{n \to \infty} \frac{1/(n+1)}{1/[2\sqrt{n+1}]} = \lim_{n \to \infty} \frac{2\sqrt{n+1}}{n+1} = \lim_{n \to \infty} \frac{2}{\sqrt{n+1}} = 0 \]Thus, through L'Hopital's Rule, we've determined that the sequence converges to 0, illustrating why it's such a powerful approach in calculus.