Math Problem Statement

The series

∑n=1∞( arctan(n+2)− arctan(n))

is a telescopic series. Calculate its sum. Give the exact value.

Solution

We are given the following telescopic series:

n=1(arctan(n+2)arctan(n))\sum_{n=1}^{\infty} \left( \arctan(n+2) - \arctan(n) \right)

Step 1: Express the series terms

This is a telescoping series, where consecutive terms cancel each other. To understand this cancellation better, let’s expand the first few terms:

  • The first term is: arctan(3)arctan(1)\arctan(3) - \arctan(1)
  • The second term is: arctan(4)arctan(2)\arctan(4) - \arctan(2)
  • The third term is: arctan(5)arctan(3)\arctan(5) - \arctan(3)
  • The fourth term is: arctan(6)arctan(4)\arctan(6) - \arctan(4)
  • And so on...

Step 2: Observe the cancellation pattern

You can see that most terms cancel out. Specifically, for every term arctan(n+2)arctan(n)\arctan(n+2) - \arctan(n), the arctan(n)\arctan(n) from one term cancels with the arctan(n+2)\arctan(n+2) from the next term.

Thus, after the cancellation, only the first few and last terms remain:

Remaining terms after cancellation: limN(arctan(N+2)arctan(1))\text{Remaining terms after cancellation: } \lim_{N \to \infty} \left( \arctan(N+2) - \arctan(1) \right)

Step 3: Take the limit

As NN approaches infinity, arctan(N+2)\arctan(N+2) approaches π2\frac{\pi}{2} because arctan(x)\arctan(x) approaches π2\frac{\pi}{2} as xx \to \infty.

Thus, the sum of the series is:

π2arctan(1)\frac{\pi}{2} - \arctan(1)

Step 4: Calculate arctan(1)\arctan(1)

We know that arctan(1)=π4\arctan(1) = \frac{\pi}{4}, since the tangent of π4\frac{\pi}{4} is 1.

Step 5: Final sum

Now, substitute this value into the expression:

π2π4=π4\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}

Thus, the sum of the series is:

π4\boxed{\frac{\pi}{4}}


Would you like to explore the telescoping property in more detail or ask any questions about the calculation? Here are some related questions to consider:

  1. What is the general form of a telescoping series?
  2. How does the cancellation process work in telescoping series?
  3. How can we apply telescoping series to other types of trigonometric functions?
  4. How does the concept of limits help in evaluating infinite series like this?
  5. Can you find a telescoping series involving other inverse trigonometric functions?

Tip: When dealing with telescoping series, it helps to write out the first few terms explicitly to understand the cancellation pattern better.

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Math Problem Analysis

Mathematical Concepts

Telescoping Series
Arctangent Function
Limits

Formulas

Sum of telescoping series
Arctan(x) formula
Limit of arctan(x) as x approaches infinity

Theorems

Telescoping series cancellation
Limit of inverse trigonometric functions

Suitable Grade Level

Grades 11-12