Question

The following series converge by the ratio test. Use summation by parts, \(\sum_{k=1}^{n} a_{k}\left(b_{k+1}-b_{k}\right)=\left[a_{n+1} b_{n+1}-a_{1} b_{1}\right]-\sum_{k=1}^{n} b_{k+1}\left(a_{k+1}-a_{k}\right)\), to find the sum of the given series. $$ \left.\sum_{k=1}^{\infty} \frac{k}{c^{k}}, \text { where } c>1 \text { (Hint: Take } a_{k}=k \text { and } b_{k}=c^{1-k} /(c-1) .\right) $$

Short answer

The sum of the series is \(\frac{1}{(c-1)^2}\).

Step-by-step solution

Understanding the Formula

We are given the summation by parts formula: \[\sum_{k=1}^{n} a_{k}(b_{k+1} - b_{k}) = [a_{n+1}b_{n+1} - a_1b_1] - \sum_{k=1}^{n} b_{k+1}(a_{k+1} - a_{k})\]We need to use this to find the sum of a series \(\sum_{k=1}^{\infty} \frac{k}{c^k}\). We need to choose suitable sequences \(a_k\) and \(b_k\).

Defining Sequences

According to the hint, set \(a_k = k\) and \(b_k = \frac{c^{1-k}}{c-1}\). This gives us:\[b_{k+1} = \frac{c^{1-(k+1)}}{c-1} = \frac{c^{-k}}{c(c-1)} = \frac{1}{c^{k+1}(c-1)}\]

Applying the Summation by Parts Formula

Substitute the defined sequences into the formula:\[\sum_{k=1}^{n} a_{k}(b_{k+1} - b_{k}) = [a_{n+1}b_{n+1} - a_1b_1] - \sum_{k=1}^{n} b_{k+1}(a_{k+1} - a_{k})\]Calculate \(b_{k+1} - b_k\) and \(a_{k+1} - a_k = 1\).

Simplifying Terms

Compute each part:1. Find \(a_{n+1}b_{n+1} = (n+1) \cdot \frac{c^{1-(n+1)}}{c-1} = \frac{n+1}{c^n(c-1)}\)2. Find \(a_1b_1 = 1 \cdot \frac{c^{1-1}}{c-1} = \frac{1}{c-1}\)3. Evaluate the sum \(\sum_{k=1}^{n} b_{k+1}(a_{k+1} - a_{k}) = \sum_{k=1}^{n} \frac{1}{c^{k+1}(c-1)} = \frac{1}{c-1}\sum_{k=1}^{n} \frac{1}{c^{k+1}}\)

Conclusion

As \(n\) approaches infinity, the term \(\frac{n+1}{c^n(c-1)}\) approaches zero since \(c > 1\). Thus:\[\lim_{n \to \infty} \left( \frac{n+1}{c^n(c-1)} - \frac{1}{c-1} \right) = -\frac{1}{c-1}\]We also recognize that the sum \( \frac{1}{c-1}\sum_{k=1}^{\infty} \frac{1}{c^{k+1}} = \frac{1}{c-1}\cdot \frac{1}{c(c-1)} = \frac{1}{c(c-1)^2} \).Thus, the series sum is \(-\frac{1}{c-1} + \frac{1}{c(c-1)^2} = \frac{1}{(c-1)^2}\).

Key concepts

Ratio Test

The ratio test is a powerful method used to determine the convergence or divergence of infinite series. It's particularly useful when dealing with series whose terms involve factorials, exponentials, or powers, as it provides a straightforward way to assess whether a series converges. The test itself revolves around the limit of the absolute value of the ratio of consecutive terms in a series. More formally, for a series \( \sum_{n=1}^{\infty} a_n \), you calculate the limit:

• \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \)

If this limit is less than 1, the series converges absolutely. If it is greater than 1, or infinite, the series diverges. If this limit equals 1, the test is inconclusive, and another method must be used to determine convergence or divergence.

In the context of the exercise provided, the ratio test helps confirm the convergence status of the series \( \sum_{k=1}^{\infty} \frac{k}{c^k} \). By evaluating the ratio of successive terms and ensuring it falls below 1, you can conclude that the ongoing sum of terms will not diverge to infinity.

Summation by Parts

Summation by parts is an analog of integration by parts for discrete functions, essentially dealing with sequences and series. This technique transforms the summation of products into a more manageable form by strategically choosing two sequences, \( a_k \) and \( b_k \). The formula used is:

• \( \sum_{k=1}^{n} a_{k}(b_{k+1} - b_{k}) = [a_{n+1}b_{n+1} - a_1b_1] - \sum_{k=1}^{n} b_{k+1}(a_{k+1} - a_{k}) \)

By cleverly choosing your sequences, you can simplify the original sum into a form that can be more easily calculated or approximated. For example, in our exercise, \( a_k = k \) and \( b_k = \frac{c^{1-k}}{c-1} \) were selected to simplify the resolution of the infinite series \( \sum_{k=1}^{\infty} \frac{k}{c^k} \).

The method allows for the decomposition of a complex expression into parts that cancel out or converge to a simpler expression as \( n \) approaches infinity, effectively enabling the series' sum evaluation.

Geometric Series

A geometric series is one where each term after the first is found by multiplying the previous term by a constant factor called the common ratio. A key property of geometric series is that, when the absolute value of the common ratio \( r \) is less than 1, the series converges to a finite sum. This sum can be calculated using the formula:

• \( \sum_{k=0}^{\infty} ar^k = \frac{a}{1-r} \)

where \( a \) is the first term.

In our exercise context, once the complex sum was decomposed using summation by parts, it included a component reflecting a geometric series \( \sum_{k=1}^{\infty} r^k \), where \( r = \frac{1}{c} \). Since \( c > 1 \), \( r < 1 \) and therefore, this series absolutely converges, making it critical in determining the final sum of the originally given series.

Convergence

Convergence is a fundamental concept in the analysis of series and sequences. In the context of a series, convergence means that as you sum more and more terms, the total approaches a fixed value, rather than continuing to grow indefinitely. Understanding whether a series converges is crucial for evaluating or decomposing its sum.

Various tests exist to determine convergence, such as the ratio test mentioned earlier. Each test provides insight into whether a series stabilizes to a finite value as its number of terms increases. In our example, by applying these methods, particularly the ratio test and the nature of geometric series, we concluded that the series \( \sum_{k=1}^{\infty} \frac{k}{c^k} \) converges to \( \frac{1}{(c-1)^2} \).

The convergence of a series is not just about finding its sum but also about understanding its behavior, ensuring that the result has practical reliability in mathematical calculations and real-world applications.