Problem 36 - Precalculus: Mathematical for Calculus

Question

Find the first four partial sums and the \(n\) th partial sum of the sequence \(a_{n}\) $$a_{n}=\frac{1}{n+1}-\frac{1}{n+2}$$

Step-by-Step Solution

Answer

First four partial sums are \(\frac{1}{6}, \frac{1}{4}, \frac{3}{10}, \frac{1}{3}\). \(n\)th partial sum is \(S_n = \frac{1}{2} - \frac{1}{n+2}\).

Step 1: Understand the sequence

The given sequence is defined as \(a_n = \frac{1}{n+1} - \frac{1}{n+2}\). Each term of the sequence is the difference between the reciprocals of consecutive integers starting from \(n+1\). We need to find the partial sums up to the fourth term and the general formula for the \(n\)th partial sum.

Step 2: Calculate the first partial sum \(S_1\)

The first partial sum \(S_1\) is simply the first term \(a_1\). Calculate \(a_1\) using the formula: \(a_1 = \frac{1}{1+1} - \frac{1}{1+2} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\). So, \(S_1 = \frac{1}{6}\).

Step 3: Calculate the second partial sum \(S_2\)

The second partial sum \(S_2\) is \(a_1 + a_2\). First, find \(a_2\):\[a_2 = \frac{1}{2+1} - \frac{1}{2+2} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}\].\Thus, \(S_2 = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}\).

Step 4: Calculate the third partial sum \(S_3\)

The third partial sum \(S_3\) is \(S_2 + a_3\). First, find \(a_3\):\[a_3 = \frac{1}{3+1} - \frac{1}{3+2} = \frac{1}{4} - \frac{1}{5} = \frac{1}{20}\].\So, \(S_3 = \frac{1}{4} + \frac{1}{20} = \frac{5}{20} + \frac{1}{20} = \frac{6}{20} = \frac{3}{10}\).

Step 5: Calculate the fourth partial sum \(S_4\)

The fourth partial sum \(S_4\) is \(S_3 + a_4\). Find \(a_4\):\[a_4 = \frac{1}{4+1} - \frac{1}{4+2} = \frac{1}{5} - \frac{1}{6} = \frac{1}{30}\].\Thus, \(S_4 = \frac{3}{10} + \frac{1}{30} = \frac{9}{30} + \frac{1}{30} = \frac{10}{30} = \frac{1}{3}\).

Step 6: Find the general formula for the \(n\)th partial sum

The sequence \(a_n\) telescopes. When you sum terms from 1 to \(n\), most intermediate terms cancel out. The \(n\)th partial sum can be derived as: \[S_n = \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \].\Thus, \(S_n = \frac{1}{2} - \frac{1}{n+2}\).

Key Concepts

Telescoping Series

A telescoping series is an interesting type of series where many terms cancel each other out, leaving only a few terms that sum up to the total value. In mathematics, recognizing a telescoping series can simplify the process of finding partial sums significantly.

In our case, the sequence given is: \[ a_n = \frac{1}{n+1} - \frac{1}{n+2} \]. Each term is the difference between consecutive fractions. The beauty of a telescoping sequence like this is seen when you arrange the terms for summation. The sequence is:

• \(a_1 = \frac{1}{2} - \frac{1}{3}\)
• \(a_2 = \frac{1}{3} - \frac{1}{4}\)
• \(a_3 = \frac{1}{4} - \frac{1}{5}\)

Each middle term like \(\frac{1}{3}\) and \(\frac{1}{4}\) cancels out with a corresponding term or terms in the series. This ultimately helps to simplify the series to just the first (\(\frac{1}{2}\)) and a remaining part (\( - \frac{1}{n+2}\)) at the end of the sequence. In this way, telescoping series are a very effective method to compute sums where significant cancellation occurs.

Sequence

A sequence is essentially a list of numbers in a specific order. Each number in the sequence is called a term. Sequence notation often uses \(a_n\) to denote the \(n\)-th term. For example, in the problem at hand, the sequence is defined as: \[ a_n = \frac{1}{n+1} - \frac{1}{n+2} \]. This means that each term of the sequence is formed by subtracting one fraction from the next one.

Understanding sequences fully involves recognizing the pattern or formula behind the numbers. In this case:

• The numerator is always 1.
• The denominators increase by 1 with each succeeding term.

Sequences are fundamental in understanding more complex mathematical concepts such as series, limits, and functions. By breaking down each term's pattern, students can often predict future terms or sum multiple terms, which is crucial for finding partial sums.

nth Partial Sum

An \(n\)-th partial sum represents the sum of the first \(n\) terms in a sequence. This concept is critical in calculus and other areas of mathematics since it provides insight into the behavior of sequences and series.

For our sequence \(a_n = \frac{1}{n+1} - \frac{1}{n+2}\), the partial sum \(S_n\) wants you to add up terms \(a_1 + a_2 + \ldots + a_n\). With a telescoping nature of this series:

• The series starts at \(\frac{1}{2}\) (as seen with \(a_1\)).
• The series eventually ends at \(-\frac{1}{n+2}\), following the natural cancellation of intermediate terms.

Thus, for the general \(n\)-th partial sum, you get a formula revealing the simplified form: \(S_n = \frac{1}{2} - \frac{1}{n+2}\). This shows how effectively partial sums represent the whole sequence's behavior up to a certain point, providing a handy tool to predict and understand further terms without expanding each one individually.