Question

a. Evaluate the series $$ \sum_{k=1}^{\infty} \frac{3^{k}}{\left(3^{k+1}-1\right)\left(3^{k}-1\right)} $$ b. For what values of \(a\) does the series $$ \sum_{k=1}^{\infty} \frac{a^{k}}{\left(a^{k+1}-1\right)\left(a^{k}-1\right)} $$ converge, and in those cases, what is its value?

Short answer

Answer: The series converges for all values of \(a > 0\), and the sum is \(-\frac{1}{a - 1}\).

Step-by-step solution

Rewrite the series as a telescoping series

To transform the series into a telescoping series, we will decompose the given term into partial fractions. This means finding two constants A and B such that: $$ \frac{a^k}{\left(a^{k+1}-1\right)\left(a^{k}-1\right)} = \frac{A}{a^{k+1} - 1} + \frac{B}{a^{k} - 1} $$ Now, we multiply both sides by \(\left(a^{k+1}-1\right)\left(a^{k}-1\right)\) to get: $$ a^k = A (a^k - 1) + B (a^{k+1} - 1) $$ We can rewrite this as: $$ a^k = A a^{k+1} - A + B a^k - B a^{k+1} $$

Solve for A and B

From the equation obtained in Step 1, we see that: $$ A a^{k+1} - B a^{k+1} + B a^k - A a^k = a^k $$ This has to be true for all values of k. For simplification, let's choose values of k such that: - When \(k=0\), we have: $$ A - B + B - A = 1 \Rightarrow B = 1 $$ - When \(k=1\), we have: $$ A a^2 - B a^2 + B a - A a = a \Rightarrow (A - B) a^2 + B a - A a = a \Rightarrow A a^2 - A a = a $$ Now, knowing that \(B=1\), we get \(A a - A = 1\), and hence \(A(a-1) = 1\), which gives us \(A = \frac{1}{a-1}\). So, we have: $$ \frac{a^k}{\left(a^{k+1}-1\right)\left(a^{k}-1\right)} = \frac{\frac{1}{a-1}}{a^{k+1} - 1} + \frac{1}{a^{k} - 1} $$

Rewrite the series and find the sum (for part a)

Now we can rewrite the series for part (a), using \(a = 3\), as: $$ \sum_{k=1}^{\infty} \frac{3^{k}}{\left(3^{k+1}-1\right)\left(3^{k}-1\right)} = \sum_{k=1}^{\infty} \left(\frac{\frac{1}{2}}{3^{k+1} - 1} + \frac{1}{3^{k} - 1}\right) $$ This sum is telescoping, which means that many of the terms will cancel each other out. Notice that: $$ -S_{n} = -\sum_{k=1}^{n} \left(\frac{\frac{1}{2}}{3^{k+1} - 1} + \frac{1}{3^{k} - 1}\right) = -\left(\frac{1}{2}\right) + \frac{1}{2} - \frac{1}{3^2 - 1} + \frac{1}{3^3 - 1} - \dots - \frac{1}{2(3^{n+1} - 1)} + \frac{1}{3^{n+1} - 1} $$ So, we have that: $$ \lim_{n \to\infty} S_{n} = -\frac{1}{2} + \frac{1}{2} = 0 $$ Therefore, the series in part (a) converges and has a sum of 0.

Analyze convergence of the series (for part b)

For part (b), let's analyze the behavior of the general series \(\sum_{k=1}^{\infty}\frac{a^{k}}{\left(a^{k+1}-1\right)\left(a^{k}-1\right)}\) for different values of a. If \(a = 1\), the series converges trivially as the sum is equal to the series in part (a). For \(a > 1\), the series contracts, that is, its terms decrease in magnitude as k increases, which implies that it converges. Finally, if \(a \in (0, 1)\), each term of the series is positive and decreasing, and the series converges. So the series converges for all \(a > 0\).

Find the sum for the convergent series (for part b)

Applying the same telescoping argument as in part (a), the sum of the series for part (b) can be expressed as: $$ S_n = -1\frac{1}{a-1} + \frac{1}{a^{n+1} - 1} $$ Then, the sum for the infinite series is found by taking the limit as n goes to infinity: $$ S = \lim_{n \to\infty} S_n = -1\frac{1}{a-1} + \frac{1}{0} = -\frac{1}{a-1} $$ Hence, the sum converges for all values of \(a\) in \((0, \infty)\) and has a sum of \(-\frac{1}{a - 1}\).

Key concepts

Telescoping Series

A telescoping series is an infinite series where most terms cancel each other out, leaving only a few terms that contribute to the final sum. This phenomenon usually occurs due to specific structural properties of the terms in the sequence. In a telescoping series, we typically rewrite the general term such that when added sequentially, each term cancels out a part of a neighbor term.
Consider the case of the series:

• The expansion and cancellation often result from rewriting terms as the difference of fractions or functions.
• The focus is on the boundary terms of the series, which appear in an uncancelled form at the start and end.

Telescoping is an effective strategy for evaluating series, simplifying the task of finding convergence and calculating sums.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express complex rational expressions as a sum of simpler fractions. This technique is crucial when dealing with series that appear complex at first glance. To decompose a fraction:

• Break down the original expression into a sum of fractions, ideally with simpler denominators.
• Determine the unknown constants by equating coefficients or using substitution.
• Reorganize the expression so it can be more easily analyzed or summed.

For example, in our series exercise, each term is transformed into a combination of two simple fractions. Once decomposed, these simpler terms make it easier to apply convergence tests or identify telescoping characteristics.

Convergence Criteria

Convergence is a fundamental concept in the study of infinite series. It refers to the condition when the sum of an infinite series approaches a finite limit. Various tests and observations help determine convergence:

• **Test for convergence**: Examine if the terms of a series decrease to zero as the series progresses.
• **Boundedness**: Check if partial sums remain bounded.
• Apply specific tests like the ratio test or alternating series test based on the nature of the series.

In the given problem, the criteria for convergence depend on the value of the parameter 'a'. When examining convergence: - For a = 3, the series converges to 0 due to telescoping. - For any a > 0, the series terms decrease and remain positive, implying convergence for these values.

Infinite Series Sum

Calculating the sum of an infinite series involves determining if the series converges and then finding its limit. This process requires utilizing the structural aspects of the series:

• By using telescoping, a rapidly converging pattern helps find the sum efficiently.
• The sum is often found through recognizing cancellations in extended series or transforming the series form.

For example, in part (a) of our exercise: - The telescopic form of the series made it clear that the sum was 0. In more complex cases, like part (b), understanding how series terms converge aids in calculating sums. Adjustments in parameter values lead to different results calculated by analyzing the altered series behavior and applying the known series formulas. This thoughtful approach highlights the finesse of evaluating infinite series sums.