Problem 39 - Calculus: One and Several Variables

Question

(a) Show that \(\sum_{h=0}^{\infty} e^{-\alpha A}\) converges for cach \(\alpha > 0\) (b) Show that \(\sum_{k=0}^{\infty} k e^{-\omega k}\) converges for \(\operatorname{cach} \alpha\) ? 0 . (c) Show that, more generally, \(\sum_{k=0}^{\infty} k^{n} e^{-\alpha k}\) converges for cach nonnegative integer \(n\) and \(\operatorname{cach} \alpha > 0\)

Step-by-Step Solution

Answer

In summary, we have shown the convergence of the given series: (a) \(\sum_{h=0}^{\infty} e^{-\alpha A}\) converges for all \(\alpha > 0\) as it is a geometric series with \(|e^{-\alpha A}| \leq 1\). (b) \(\sum_{k=0}^{\infty} k e^{-\omega k}\) converges for all \(\omega > 0\) using the comparison test with the sum of \(e^{-\omega k}\) from part (a). (c) \(\sum_{k=0}^{\infty} k^{n} e^{-\alpha k}\) converges for all nonnegative integers \(n\) and \(\alpha>0\) by induction and the comparison test.

Step 1: The sum of \(e^{-\alpha A}\)

Let's analyze the series given in part (a):\(\sum_{h=0}^{\infty} e^{-\alpha A}\) Notice that this sum is a geometric series with the common ratio \(e^{-\alpha A}\). The geometric series converges if \(|e^{-\alpha A}|<1\). Since \(A\geq0\) and \(\alpha>0\), \(|e^{-\alpha A}|\leq1\). Therefore, the series converges for all \(\alpha > 0\).

Step 2: The sum of \(k e^{-\omega k}\)

Now let's look at part (b) and analyze the given series: \(\sum_{k=0}^{\infty} k e^{-\omega k}\) To ensure convergence, we can use the comparison test: \[ 0 \leq k e^{-\omega k} \leq e^{-\omega k} \] We know that the sum \(\sum_{k=0}^{\infty} e^{-\omega k}\) converges from part (a), so according to the comparison test, the series \(\sum_{k=0}^{\infty} k e^{-\omega k}\) converges for all \(\omega > 0\).

Step 3: The sum of \(k^n e^{-\alpha k}\) for nonnegative integer \(n\)

Finally, let's analyze part (c):\(\sum_{k=0}^{\infty} k^{n} e^{-\alpha k}\), where \(n\) is a nonnegative integer and \(\alpha > 0\). We will use induction on \(n\). We have already shown the base case \(n=1\) in part (b). Now, let's assume that the series \(\sum_{k=0}^{\infty} k^{n-1} e^{-\alpha k}\) converges for some nonnegative integer \(n-1\) and \(\alpha>0\). We want to show that the series \(\sum_{k=0}^{\infty} k^n e^{-\alpha k}\) converges as well. As we did for part (b), we can use the comparison test to show that the series \(\sum_{k=0}^{\infty} k^n e^{-\alpha k}\) converges. Since for all \(k>2\), \[ 0 \leq \frac{k!}{(k-n)!} e^{-\alpha k} \leq k^n e^{-\alpha k} \] Note that the left side converges (\(k! e^{-\alpha k}\)) as shown by Stirling's approximation and the comparison test with the sum of \(e^{-\alpha k}\). Therefore, by the comparison test, the sum \(\sum_{k=0}^{\infty} k^n e^{-\alpha k}\) converges for all nonnegative integers \(n\) and \(\alpha > 0\).

Key Concepts

Geometric Series

Geometric series are one of the foundational concepts in mathematics, particularly useful in identifying patterns in summations. The geometric series has a characteristic form:

• The first term of the series, often denoted as \(a\).
• A common ratio, \(r\), which is multiplied by the previous term to find the next.

A typical geometric series looks like this: \(a + ar + ar^2 + ar^3 + \ldots\). For such a series to converge to a finite value, the absolute value of the common ratio, \(r\), must be less than 1, that is, \(|r| < 1\).
In the exercise provided, the given series \(\sum_{h=0}^{\infty} e^{-\alpha A}\) is geometric, as each term is a multiple of \(e^{-\alpha A}\). Since \(\alpha > 0\), the expression \(e^{-\alpha A}\) results in a common ratio less than 1, ensuring the series converges.

Comparison Test

The comparison test is a handy tool for determining the convergence of infinite series, especially when the series itself is complex. This method relies on comparing the series in question with another series that is already known to converge or diverge.
Here's how it works:

• If \(0 \leq a_n \leq b_n\) for all \(n\), and \(\sum b_n\) converges, then \(\sum a_n\) also converges.
• Conversely, if \(a_n \geq b_n \geq 0\) for all \(n\), and \(\sum b_n\) diverges, then \(\sum a_n\) diverges too.

In the problem, the series \(\sum_{k=0}^{\infty} k e^{-\omega k}\) is compared to \(\sum_{k=0}^{\infty} e^{-\omega k}\), a series already proven to converge for \(\omega > 0\). As \(0 \leq k e^{-\omega k} \leq e^{-\omega k}\), the series under investigation also converges.

Induction Method

Induction is a powerful technique used in mathematics to prove statements for all natural numbers. It usually involves two main steps:

• Base Case: Verify that the property holds for an initial natural number, usually \(n=1\).
• Inductive Step: Assume the property holds for \(n=k\) and then prove it for \(n=k+1\).

By following these steps, a conclusion can be extended from the base case throughout all natural numbers. In part (c) of the exercise, this method is used to show the series \(\sum_{k=0}^{\infty} k^n e^{-\alpha k}\) converges for all nonnegative integers \(n\).
The base case is checked for \(n=1\) in part (b). Then, assuming convergence for \(n-1\), the series is shown to also converge for \(n\) by using similar techniques along with a comparison test.

Stirling's Approximation

Stirling's Approximation is a formula used to approximate large factorials. Its classic form is:\[ n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \]This approximation becomes more accurate as \(n\) increases, which is particularly useful when dealing with series where factorial terms are involved and precise values are hard to compute.
In the exercise, this approximation aids in evaluating the convergence of series like \(\sum_{k=0}^{\infty} k^n e^{-\alpha k}\).
By using Stirling's approximation, proving convergence often becomes simpler as it allows certain terms to be bounded, making application of the comparison test feasible. This is crucial for determining that the exponential decay in \(e^{-\alpha k}\) can offset the growth of \(k^n\) as \(k\) becomes larger.