Math Problem Statement

In a telexcoping series, I can't find the nth term. Teach me.

Solution

In a telescoping series, the series is often expressed as a sum of terms where many terms cancel out, leaving only a few remaining terms when simplified. These series are typically written in a form that allows the terms to "collapse" or "telescope" when summed.

General Form of a Telescoping Series

A telescoping series has the following form:

Sn=k=1n(akak+1)S_n = \sum_{k=1}^n (a_k - a_{k+1})

Where:

  • aka_k and ak+1a_{k+1} are expressions that depend on the index kk.
  • As you sum the terms, most of them cancel out, and you are left with just the first and last terms.

Steps to Identify the nth Term

  1. Look for the pattern in the terms. Telescoping series are generally structured as differences between consecutive terms, like akak+1a_k - a_{k+1}.

  2. Write out a few terms in the series to observe the cancellation. For example: Sn=(a1a2)+(a2a3)+(a3a4)++(anan+1)S_n = (a_1 - a_2) + (a_2 - a_3) + (a_3 - a_4) + \dots + (a_n - a_{n+1})

    You'll notice that every intermediate term cancels out, and you're left with: Sn=a1an+1S_n = a_1 - a_{n+1}

  3. Find the general form for aka_k. This typically involves looking at the given series' terms and recognizing the form of aka_k.

  4. Sum up the series by evaluating the limits of the remaining terms.

Example 1: Simple Telescoping Series

Consider the series:

Sn=k=1n(1k1k+1)S_n = \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right)

Let’s break it down step-by-step:

  • For k=1k = 1, the term is 1112\frac{1}{1} - \frac{1}{2}.
  • For k=2k = 2, the term is 1213\frac{1}{2} - \frac{1}{3}.
  • For k=3k = 3, the term is 1314\frac{1}{3} - \frac{1}{4}, and so on.

When you sum the series:

Sn=(1112)+(1213)+(1314)++(1n1n+1)S_n = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{n} - \frac{1}{n+1} \right)

The terms cancel in the middle, leaving you with:

Sn=11n+1S_n = 1 - \frac{1}{n+1}

Thus, the nth term can be expressed as:

Sn=11n+1S_n = 1 - \frac{1}{n+1}

Key Steps in Finding nth Term for Telescoping Series:

  • Recognize the general structure of the series.
  • Identify the cancellation pattern and write out terms to see how they telescope.
  • Simplify the sum, and use the first and last terms to express the final result.

Would you like to see more examples, or are there any specific questions you have about telescoping series? Here are some further topics you might be interested in:

  1. How to handle telescoping series involving polynomials or other functions.
  2. The convergence of telescoping series.
  3. Deriving the nth term for more complex telescoping series.
  4. Identifying telescoping series in mathematical problems.
  5. Applying telescoping series in integration and summation.

Tip: Always try to write a few terms out explicitly to visualize how the cancellation works.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Series and Sequences
Telescoping Series
Summation

Formulas

S_n = ∑ (a_k - a_{k+1})
S_n = a_1 - a_{n+1}

Theorems

Summation of Telescoping Series

Suitable Grade Level

Grades 10-12