Question

For the following telescoping series, find a formula for the nth term of the sequence of partial sums \(\left\\{S_{n}\right\\} .\) Then evaluate lim \(S_{n}\) to obtain the value of the series or state that the series diverges. \(^{n \rightarrow \infty}\) $$\sum_{k=0}^{\infty} \frac{1}{(3 k+1)(3 k+4)}$$

Short answer

Answer: The value of the series is $\frac{1}{8}$.

Step-by-step solution

Rewrite the general term k

To start, let's rewrite the given sum in terms of k: $$\frac{1}{(3 k+1)(3 k+4)}$$ We will try to decompose it into partial fractions: $$\frac{1}{(3 k+1)(3 k+4)}=\frac{A}{3 k+1} + \frac{B}{3 k+4}$$ Where A and B are constants we need to find.

Determine the constants A and B

To determine values of A and B, we can clear denominators by multiplying both sides by \((3 k+1)(3 k+4)\): $$1=A(3 k+4)+B(3 k+1)$$ For k = 0, we get: $$1=4A \Rightarrow A=\frac{1}{4}$$ For k = -1, we get: $$1 = -2B \Rightarrow B=-\frac{1}{2}$$ Now we can rewrite our sum with the determined values of A and B: $$\frac{1}{(3 k+1)(3 k+4)}=\frac{1/4}{3 k+1} + \frac{-1/2}{3 k+4}$$

Write the partial sum formula

Let's find the formula for the sequence of partial sums \(S_n\): $$S_{n}=\sum_{k=0}^{n} \left\{ \frac{1/4}{3 k+1} + \frac{-1/2}{3 k+4} \right\}$$ As the sums telescopes, we can write it as: $$S_{n}=\frac{1}{4}\left(\frac{1}{1}-\frac{1}{1+3}+\frac{1}{3+1}-\frac{1}{3+1+3}+\dots+\frac{1}{3(n-1)+1}-\frac{1}{3(n-1)+1+3}\right)-\frac{1}{2}\left(\frac{1}{4}-\frac{1}{4+3}+\frac{1}{3+4}-\frac{1}{3+4+3}+\dots+\frac{1}{3n+1}-\frac{1}{3n+1+3}\right)$$ This simplifies to: $$S_{n}=\frac{1}{4}\left(1-\frac{1}{3n+4}\right)-\frac{1}{2}\left(\frac{1}{4}-\frac{1}{3n+7}\right)$$

Find the limit of the partial sum

Now, we will find the limit of the partial sum as n approaches infinity to check the convergence of the series: $$\lim_{n\to\infty} S_{n}=\lim_{n\to\infty} \left[\frac{1}{4}\left(1-\frac{1}{3n+4}\right)-\frac{1}{2}\left(\frac{1}{4}-\frac{1}{3n+7}\right)\right]$$ Since the limit of (\(1/(3n+4)\)) and (\(1/(3n+7)\)) as n approaches infinity is 0, we get: $$\lim_{n\to\infty} S_{n} = \frac{1}{4}(1) - \frac{1}{2}\left(\frac{1}{4}\right) = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}$$ The limit exists and is equal to \(\frac{1}{8}\). Therefore, the series converges, and the value of the series is: $$\sum_{k=0}^{\infty} \frac{1}{(3 k+1)(3 k+4)}=\frac{1}{8}$$

Key concepts

Partial Fractions

Partial fractions play a key role in simplifying complex rational expressions. When you have a fraction with a polynomial in the denominator, like \( \frac{1}{(3k+1)(3k+4)} \), you can't easily perform operations with it, such as integration or finding partial sums. This is where partial fraction decomposition comes in handy.
This process involves breaking down a complicated fraction into a sum of simpler fractions. For a partial fraction decomposition, we express:

• \( \frac{1}{(3k+1)(3k+4)} \)
• as \( \frac{A}{3k+1} + \frac{B}{3k+4} \)

Using algebra, we need to determine the constants \(A\) and \(B\). By clearing the denominators and matching coefficients, we find suitable values for these constants that satisfy the original equation. In the example, \(A = \frac{1}{4}\) and \(B = -\frac{1}{2}\), allowing for a simplified computation of sums over the series. This makes the math manageable and opens the door for telescoping, greatly simplifying subsequent steps.

Convergence of Series

Understanding the convergence or divergence of a series is crucial in determining if the series adds to a specific, finite value.For the series \(\sum_{k=0}^{\infty} \frac{1}{(3 k+1)(3 k+4)}\), we're interested in finding if the infinite sum results in a finite number or not. One way to approach this is by examining the partial sums \(S_n\), defined as the sum of the first \(n\) terms of the series.
If the sequence of partial sums \(\{S_n\}\) converges to a limit as \(n\) approaches infinity, then the series itself converges to this limit. In our problem, we use this approach to find if the series converges.
After simplifying the series with partial fractions, we observe the series undergoes telescoping, where many terms cancel. This kind of simplification helps reveal the convergence properties. Ultimately, the limit \(\lim_{n\to\infty} S_n\) will tell us if the sum is finite. If it reaches a finite value, the series is convergent, as we confirm with the value \( \frac{1}{8} \).

Partial Sums

Partial sums, denoted usually by \(S_n\), form the basis of summing infinite series incrementally. For the series in question, the partial sum \(S_n\) is defined as:

• \( S_n = \sum_{k=0}^{n} \left( \frac{1/4}{3k+1} + \frac{-1/2}{3k+4} \right) \)

This expression shows how partial sums are constructed from the series' terms up to the \(n\)-th term. The telescoping nature of this series means that many terms cancel each other out when computed for consecutive \(n\), simplifying the expression for \(S_n\).
With simplification, \(S_n\) becomes:

• \( S_{n}=\frac{1}{4}\left(1-\frac{1}{3n+4}\right)-\frac{1}{2}\left(\frac{1}{4}-\frac{1}{3n+7}\right) \)

These partial sums give us insights into the overall behavior of the series. Evaluating the limit of \(S_n\) as \(n\) grows helps us determine the convergence of the series.

Limits of Sequences

Calculating the limit of a sequence of partial sums gives us critical information about the convergence of an entire series. This is the last, essential step, where we examine the behavior of \(\lim_{n\to\infty} S_n\).
In the example provided, we find the limit of:

• \(\frac{1}{4}(1-\frac{1}{3n+4})-\frac{1}{2}(\frac{1}{4}-\frac{1}{3n+7})\)

As \(n\) approaches infinity, certain components \(\left(\frac{1}{3n+4}\) and \(\frac{1}{3n+7}\right)\) go to zero. This allows us to focus on constants in the expression, leading to a finite limit.
In our case, the limiting process results in \(\frac{1}{4} - \frac{1}{8} = \frac{1}{8}\), indicating that the series converges to this value. By finding this limit, we learn not only about the sum of the series but also verify the convergence. It reassures us that even though an infinite number of terms are involved, they summarize to a finite and predictable value.