Problem 18 - A Course in Calculus and Real Analysis

Question

Let \(\sum_{k=1}^{\infty} b_{k}\) be a convergent series. Show that the series $$ \sum_{k=1}^{\infty} k^{1 / k} b_{k} \quad \text { and } \quad \sum_{k=1}^{\infty}\left(1+\frac{1}{k}\right)^{k} b_{k} $$ are also convergent. (Hint: Exercise 17 and Exercises 7,8 of Chapter 2.)

Step-by-Step Solution

Answer

By comparing the given series with terms of known convergent series and using the Comparison Test, we found that both series \(\sum_{k=1}^{\infty} k^{\frac{1}{k}} b_{k}\) and \(\sum_{k=1}^{\infty} \left(1+\frac{1}{k}\right)^{k} b_{k}\) are convergent. This was achieved by showing that their absolute value terms are respectively bounded by \(2|b_{k}|\) and \(e|b_{k}|\), for which their corresponding series are convergent.

Step 1: Find absolute value of terms for series 2 and series 3

First, let's find the absolute value of terms for series 2 and series 3, which are: 1. \(|k^{\frac{1}{k}}b_{k}|\) 2. \(|\left(1+\frac{1}{k}\right)^{k} b_{k}|\)

Step 2: Compare with convergent series terms (Comparison Test)

Now, we need to compare the terms of series 2 and series 3 (absolute value) with the terms of the known convergent series. 1. For the series \(\sum_{k=1}^{\infty} k^{\frac{1}{k}} b_{k}\), we have that \(k^{\frac{1}{k}} \leq 2\) (from Exercise 17). Hence, \(|k^{\frac{1}{k}} b_{k}| \leq 2 |b_{k}|\). Since \(\sum_{k=1}^{\infty} b_{k}\) is given to be convergent, \(\sum_{k=1}^{\infty} 2|b_{k}|\) is also convergent. By the Comparison test, \(\sum_{k=1}^{\infty} k^{\frac{1}{k}} b_{k}\) is convergent. 2. For the series \(\sum_{k=1}^{\infty} \left(1+\frac{1}{k}\right)^{k} b_{k}\), we have that \(\left(1+\frac{1}{k}\right)^{k} < e\) (from Exercises 7 and 8 of Chapter 2). Hence, \(|\left(1+\frac{1}{k}\right)^{k} b_{k}| < e |b_{k}|\). Since \(\sum_{k=1}^{\infty} b_{k}\) is given to be convergent, \(\sum_{k=1}^{\infty} e|b_{k}|\) is also convergent. By the Comparison test, \(\sum_{k=1}^{\infty} \left(1+\frac{1}{k}\right)^{k} b_{k}\) is convergent. Hence, both the series \(\sum_{k=1}^{\infty} k^{\frac{1}{k}} b_{k}\) and \(\sum_{k=1}^{\infty} \left(1+\frac{1}{k}\right)^{k} b_{k}\) are convergent.

Key Concepts

Comparison Test

The comparison test is a useful tool for determining the convergence of a series. It works by comparing the series in question with another series whose behavior is already understood.

• If both series have positive terms and you can show that each term of the series in question is less than or equal to the corresponding term of a known convergent series, then the series in question is also convergent.
• On the other hand, if each term is greater than or equal to a corresponding term in a divergent series, then the series in question is divergent as well.

For example, in our problem, since we know that \( \sum_{k=1}^{\infty} b_k \) is convergent, we can use this fact to compare with \( |k^{1/k} b_k| \) and \( |(1+1/k)^k b_k| \). By showing these terms are less than multiples of the converging sequence, the comparison test confirms the convergence of these series.

Absolute Convergence

When discussing series convergence, one might encounter absolute convergence. This is a stronger form of convergence where a series \(\sum_{k=1}^{\infty} a_k\) converges absolutely if \(\sum_{k=1}^{\infty} |a_k|\) is convergent.

• If a series is absolutely convergent, it implies that the series is also convergent.
• Absolute convergence considers the sum of the absolute values of the terms, allowing us to ignore any negative values which may cancel out positive ones in typical convergence.

In our exercise, by showing that the absolute values \( |k^{1/k} b_k| \) and \( |(1+1/k)^k b_k| \) are bounded by a multiple of a convergent series, absolute convergence can assure us of the overall convergence, regardless of the signs of the terms.

Factorial Growth

Factorial growth exhibits a rapid increase in the size of numbers due to the nature of the factorial operation.

• Factorials are represented by the notation \( n! \,\), which is equal to \( n \times (n-1) \times \ldots \times 1 \,\) and results in very large numbers as \( n \,\) increases.
• In series, factorials can influence the rate of growth significantly, often making terms rapidly large, sometimes too large for convergence.

While factorial growth is not directly a part of the given series, understanding it highlights why many series include terms like \( k^{1/k} \,\) which grow much slower compared to factorial terms when \( n \,\) is large. This characteristic is crucial in ensuring the series converges.

Exponential Growth

Exponential growth is a pattern where numbers increase multiplicatively, sharply contrasting with arithmetic or linear growth.

• Exponential growth for a base can be demonstrated by the expression \( a^n, \,\) where \( a \,\) is a constant and the exponent \( n \,\) grows incrementally, resulting in a rapid escalation.
• In the context of series, terms that grow exponentially may affect convergence, but when controlled by a convergent factor like \( b_k \,\), they might not prevent overall convergence.

The exercise hints at this through terms like \( (1+1/k)^k \,\), approximating \( e \,\) and showing how such terms controlled within a series could still lead to convergence when managed by the right comparative methods.