Question

Suppose that \(f(x) \geq 0\) for \(x \geq k .\) Prove that \(\int_{k}^{\infty} f(x) d x<\infty\) if and only if $$ \sum_{n=k}^{\infty} \int_{n}^{n+1} f(x) d x<\infty $$ HINT: Use Theorems 3.4 .5 and \(4.3 .8 .\)

Short answer

Question: Prove that for a non-negative function f(x) ≥ 0 for x ≥ k, the integral of f(x) from k to infinity converges if and only if the series of the integral of f(x) from n to n+1 for n = k, k+1, k+2, ... converges. Use theorems 3.4.5 and 4.3.8 in your proof. Answer: We have shown that if the given sum converges, the integral converges and vice versa, using Theorem 3.4.5 (properties of non-negative functions on an interval) and Theorem 4.3.8 (comparison test for infinite series and integrals). Therefore, we conclude that the integral of f(x) from k to infinity converges if and only if the series of the integral of f(x) from n to n+1 for n = k, k+1, k+2, ... converges.

Step-by-step solution

Understand the given condition

We are given that the function \(f(x) \geq 0\) for \(x \geq k\). This means that the function is non-negative for all values of x bigger than or equal to k.

Set up the integrals and series

We need to prove that: $$ \int_{k}^{\infty} f(x) dx < \infty \Leftrightarrow \sum_{n=k}^{\infty} \int_{n}^{n+1} f(x) dx < \infty $$ meaning if one of them is true, the other one is also true.

Use Theorem 3.4.5

According to Theorem 3.4.5, for any non-negative function \(f(x)\) on an interval \([a, b]\): $$ \int_{a}^{b} f(x) dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} f(a + i \tilde{h}) \tilde{h} $$ where \(\tilde{h} = \frac{b-a}{n}\). In our case, we have the non-negative function \(f(x)\), and the interval is \([k, \infty)\) for the integral and \([n, n+1]\) for the sum. We can use this theorem for both conditions to express them in terms of each other.

Use Theorem 4.3.8

According to Theorem 4.3.8 (comparison test for infinite series and integrals), if the sum (series) converges, then the integral also converges or vice versa. Now, consider the sum: $$ \sum_{n=k}^{\infty} \int_{n}^{n+1} f(x) dx $$ Applying Theorem 3.4.5, the series is a finite sum since it converges. According to the comparison test (Theorem 4.3.8), the integral must also converge: $$ \int_{k}^{\infty} f(x) dx < \infty $$

Prove the converse

Conversely, if the integral converges: $$ \int_{k}^{\infty} f(x) dx < \infty $$ Applying Theorem 3.4.5, we can express the integral in terms of the sum: $$ \lim_{n \to \infty} \sum_{i=0}^{n-1} f(k + i \tilde{h}) \tilde{h} $$ where \(\tilde{h} = \frac{\infty-k}{n}\). Since the integral converges, the sum must also converge. According to the comparison test (Theorem 4.3.8), this implies that the given series converges too: $$ \sum_{n=k}^{\infty} \int_{n}^{n+1} f(x) dx < \infty $$

Conclusion

We have shown that if the given sum converges, the integral converges and vice versa, using Theorem 3.4.5 and Theorem 4.3.8. Therefore, we conclude that: $$ \int_{k}^{\infty} f(x) dx < \infty \Leftrightarrow \sum_{n=k}^{\infty} \int_{n}^{n+1} f(x) dx < \infty $$

Key concepts

Convergence of Integrals

When dealing with integrals, especially those stretching to infinity, it is important to determine if they converge or not. A convergent integral is one where as you keep adding up the values, the total approaches a finite number. This is like filling a bucket where water is poured in gradually; if the bucket gets full and stable after a while, it has reached convergence.

In mathematical terms, for an integral from a given point to infinity, convergence means that the area under the curve, when extended indefinitely, sums up to a fixed finite value. Here, we're considering the integral \( \int_{k}^{\infty} f(x) \, dx \). This converges if adding up the areas from \( k \) to larger and larger numbers results in a finite number.

It's essential in problems where you need to know if the function has a bound, to determine the behavior of functions over infinite intervals and ensure they don't diverge (grow towards infinity). This can help forecast real-world phenomenon like population growth or financial models over time.

Non-negative Functions

A non-negative function is one that does not dip below the horizontal axis in the graph. This means its value is zero or positive for all inputs within a specific range, usually written as \( f(x) \geq 0 \).

Because our function \( f(x) \) is non-negative from \( x = k \) onwards, we are guaranteed that the integration or summation of areas is always building up; never losing value by going negative. This is like stacking blocks that only contribute to height - never dismantling them.

Understanding that \( f(x) \geq 0 \) ensures that the behavior of the integral or series in our theorems is consistent and predictable, making it easier to apply rules regarding convergence. This property plays a key role in proving that certain calculus operations are valid and in demonstrating the relationship between the integral and the series.

Comparison Test

The Comparison Test is a tool used to determine the convergence or divergence of a series or an integral. It relies on comparing your function or series with another function or series that is already known to be convergent or divergent.

For example, if you have known that an integral \( \int_{a}^{b} g(x) \, dx \) converges, and your function \( f(x) \leq g(x) \) for all \( x \) in your interval, then \( \int_{a}^{b} f(x) \, dx \) must also converge. Visualize this like comparing heights; if one height stays below another known stable height, it too will remain under control.

In the exercise, the Comparison Test helps establish that if the series \( \sum_{n=k}^{\infty} \int_{n}^{n+1} f(x) \, dx \) or the integral \( \int_{k}^{\infty} f(x) \, dx \) converges, then the other one does too. This double check is powerful for ensuring that both approaches to calculating the behavior of \( f(x) \) from \( k \to \infty \) agree.