Question

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.\(^{n \rightarrow \infty}\) $$\sum_{k=1}^{\infty} \frac{1}{(k+6)(k+7)}$$

Short answer

Answer: The series converges to \(\frac{1}{7}\).

Step-by-step solution

Simplify the series using partial fractions

Let's first examine the series $$\sum_{k=1}^{\infty} \frac{1}{(k+6)(k+7)}$$ We can simplify this series using partial fractions by expressing the term \(\frac{1}{(k+6)(k+7)}\) as the sum of two simpler fractions. Recall that for partial fractions, \(\frac{A}{k+6} + \frac{B}{k+7} = \frac{1}{(k+6)(k+7)}\). Multiplying both sides by the denominator, \((k+6)(k+7)\), we get: $$A(k+7) + B(k+6) = 1$$ By comparing coefficients, we can determine the values for A and B. Let \(k=-7\), then: $$A(-7+7) + B(-7+6) = -7B=1 \implies B=-\frac{1}{7}$$ Now, let \(k=-6\), then: $$A(-6+7) + B(-6+6) = A = 1$$ So our sum now becomes $$\sum_{k=1}^{\infty} \left( \frac{1}{k+6} - \frac{1}{7(k+7)} \right)$$

Calculate the nth partial sum

Now, let's calculate the nth partial sum \(S_n\): $$S_n = \sum_{k=1}^{n} \left( \frac{1}{k+6} - \frac{1}{7(k+7)} \right)$$ $$S_n = \left( \frac{1}{7} - \frac{1}{14} \right) + \left( \frac{1}{8} - \frac{1}{15} \right) + \cdots + \left( \frac{1}{n+6} - \frac{1}{7(n+7)} \right)$$ After summing, we get: $$S_n = \frac{1}{7} - \frac{1}{14} + \frac{1}{14} - \frac{1}{21} + \cdots + \frac{1}{n+6} - \frac{1}{7(n+7)}$$ The terms will cancel each other out, and we will be left with: $$S_n = \frac{1}{7} - \frac{1}{7(n+7)}$$

Evaluate the limit as n goes to infinity

Now, we'll determine the limit as n approaches infinity: $$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{1}{7} - \frac{1}{7(n+7)} \right)$$ As n goes to infinity, the term \(\frac{1}{7(n+7)}\) approaches 0, and we are left with: $$\lim_{n \to \infty} S_n = \frac{1}{7}$$ This means the series converges to \(\frac{1}{7}\).

Key concepts

Partial Fractions

Partial fractions are a crucial technique in calculus used to simplify complex rational expressions. By breaking down a single, cumbersome fraction into simpler components, calculations and integrations become manageable. For instance, in our series \(\frac{1}{(k+6)(k+7)}\), partial fractions help us reform it into two separate fractions. This breakdown allows us to analyze and manipulate the series more easily.

Here's the foundational idea: given a fraction like \(\frac{1}{(k+6)(k+7)}\), we can express it as \(\frac{A}{k+6} + \frac{B}{k+7}\). To find constants \(A\) and \(B\), multiply through by the common denominator, \((k+6)(k+7)\), resulting in:

• \(A(k+7) + B(k+6) = 1\)

Solving for \(A\) and \(B\) using suitable values for \(k\), like \(k = -7\) and \(k = -6\), we obtain \(A = 1\) and \(B = -\frac{1}{7}\). This simplifies our equation into the desired partial fractions form, making complex summations or integrations more straightforward.

Nth Partial Sum

The nth partial sum is a fundamental concept when dealing with infinite series. It helps us determine the cumulative value of the first \(n\) terms of a series. Calculating this sum gives insight into the behavior of the series as a whole.

In our example, the expression for the nth partial sum derived is:

• \(S_n = \sum_{k=1}^{n} \left( \frac{1}{k+6} - \frac{1}{7(k+7)} \right)\).

As you compute each element, you'll notice many terms in the sequence cancel out. This phenomenon is common in telescoping series, where intermediate terms negate each other, simplifying the sum considerably.

This results in a final reduced form:

• \(S_n = \frac{1}{7} - \frac{1}{7(n+7)}\).

Observing the behavior of \(S_n\) as \(n\) grows sparks clues about the series' convergence or divergence.

Limit of a Sequence

The limit of a sequence is a vital tool in understanding the behavior of series as they extend towards infinity. It indicates whether a sequence approaches a specific value, diverges to infinity, or remains indeterminate.

For our task, we've calculated the nth partial sum \(S_n\), and the challenge is finding \(\lim_{n \to \infty} S_n\). This involves evaluating:

• \(\lim_{n \to \infty} \left( \frac{1}{7} - \frac{1}{7(n+7)} \right)\).

As \(n\) grows large, \(\frac{1}{7(n+7)}\) trends towards zero because it's a fraction with a denominator increasing indefinitely. Thus, the sequence \(S_n\) converges to \(\frac{1}{7}\), meaning the original series sums to \(\frac{1}{7}\), confirming its convergence.

Recognizing these limits aids in predicting the long-term behavior of a series and verifies the process of series convergence, a core principle in calculus and mathematical analysis.