Question

Suppose that \(g, g^{\prime}\), and \(\left(g^{\prime}\right)^{2}-g g^{\prime \prime}\) are all positive on \([R, \infty)\). Show that $$ \sum \frac{g^{\prime}(n)}{g(n)}<\infty $$ if and only if \(\lim _{x \rightarrow \infty} g(x)<\infty\)

Short answer

Question: Prove that the series \(\sum \frac{g^{\prime}(n)}{g(n)}\) converges if and only if \(\lim_{x \rightarrow \infty} g(x)\) converges, given that the function \(g\), its derivative \(g'\), and \((g^{\prime})^{2}-g g^{\prime\prime}\) are all positive on \([R, \infty)\). Answer: We have shown that the series \(\sum \frac{g^{\prime}(n)}{g(n)}\) converges if and only if the limit \(\lim_{x \rightarrow \infty} g(x)\) is finite (i.e., converges). The proof is based on using the comparison test after deducing inequalities from the given conditions and applying L'Hôpital's Rule.

Step-by-step solution

Examine the given function properties

Given that \(g, g^{\prime}\), and \(\left(g^{\prime}\right)^{2}-g g^{\prime\prime}\) are all positive on \([R, \infty)\), it indicates that the function \(g\) is monotonically increasing on \([R, \infty)\). Additionally, it implies that the denominator \(g(x)\) will never be zero on the given interval.

Take the limit of the series elements

Define the series elements \(a_n = \frac{g^{\prime}(n)}{g(n)}\). To determine convergence of the series, we can apply the limit comparison test. First, we need to find the limit of \(a_n\) as \(n\) approaches infinity. $$ \lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} \frac{g^{\prime}(n)}{g(n)} $$

Use L'Hôpital's Rule

To evaluate the limit, we apply L'Hôpital's Rule since the limit is of the form \(\frac{\infty}{\infty}\). $$ \lim_{n \rightarrow \infty} \frac{g^{\prime}(n)}{g(n)} = \lim_{n \rightarrow \infty} \frac{g^{\prime\prime}(n)}{g^{\prime}(n)} $$

Apply the given conditions

Now, we have the following inequality given: $$ (g^{\prime})^{2}-g g^{\prime\prime} > 0 \Rightarrow (g^{\prime})^{2} > g g^{\prime\prime} $$ From this inequality, we can deduce the following: $$ \frac{g^{\prime \prime}(n)}{g^{\prime}(n)} < \frac{(g^{\prime}(n))^2}{g(n)} $$ Hence, the limit of the ratio of \(g^{\prime \prime}(n)\) and \(g^{\prime}(n)\) is less than the limit of the ratio of \((g^{\prime}(n))^2\) and \(g(n)\).

Determine convergence using the comparison test

If \(\lim_{n \rightarrow \infty} g(x)\) is finite, then $$ \sum \frac{(g^{\prime}(n))^2}{g(n)} $$ will also converge by the comparison test since both limits are finite. This implies that: $$ \sum \frac{g^{\prime}(n)}{g(n)} < \infty $$ On the other hand, if \(\lim_{n \rightarrow \infty} g(x)\) is infinite, then the limit of the ratio of \((g^{\prime}(n))^2\) and \(g(n)\) will also be infinite, and therefore, the series will diverge.

Final Conclusion

As a result, we have shown that the series \(\sum \frac{g^{\prime}(n)}{g(n)}\) converges if and only if \(\lim_{x \rightarrow \infty} g(x)\) is finite (i.e., converges).

Key concepts

Monotonically Increasing Function

A function is called monotonically increasing if it consistently increases or remains constant as the input value increases. This is an essential property in calculus and analysis.
It simplifies the behavior of functions, making it easier to predict their outputs. To determine if a function is monotonically increasing, look at its derivative:

• If the derivative, noted as \(g'(x)\), is positive over an interval, the function is rising on that interval.
• If \(g'(x) = 0\), the function is constant over that segment.

In the provided exercise, since \(g, g',\) and \((g')^2 - gg"\) are all positive, \(g\) is monotonically increasing on \([R, \infty)\).
This property ensures that at no point within this interval does \(g(x)\) decrease, providing a crucial foundation for analyzing the given series.

Limit Comparison Test

The limit comparison test is a powerful tool to determine convergence or divergence of series by comparing them to simpler well-known series.
It involves taking the limit of the ratio of the terms of two series as they approach infinity. For two series with terms \(a_n\) and \(b_n\):

• If \(\lim_{n \to \infty} \frac{a_n}{b_n} = c\) (where \(c > 0\) is finite), then both series converge or diverge together.
• If \(\lim_{n \to \infty} \frac{a_n}{b_n} = 0\) and \(\sum b_n\) converges, so does \(\sum a_n\).
• If \(\lim_{n \to \infty} \frac{a_n}{b_n} = \infty\) and \(\sum b_n\) diverges, so does \(\sum a_n\).

In the exercise, \(a_n = \frac{g'(n)}{g(n)}\) is compared with another series using this test. By carefully choosing a known convergent series to compare against, such as one derived from integrating similar functions, a conclusion about the convergence of the original series can be drawn.

L'Hôpital's Rule

L'Hôpital's Rule is a strategy used to find limits of indeterminate forms, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), by differentiating their numerator and denominator.
It simplifies the evaluation of limits and provides a method to approach otherwise complex calculations. Here's how it works:

• If \(\lim_{x \to a} f(x) = 0\) and \(\lim_{x \to a} g(x) = 0\) (or both go to infinity) and \(g'(x) eq 0\) near \(a\), then \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\).
• This rule can be applied repeatedly until the limit is in a determinate form or non-indeterminate.

In the solution, L'Hôpital's Rule was used to find the limit of \(\frac{g'(n)}{g(n)}\) as \(n\) approaches infinity by differentiating numerator and denominator, thereby aiding in the convergence analysis of the series.