Problem 42 - Calculus

Question

Find a nice formula for the sum $$ \begin{array}{c} \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)} \\ \text { Hint: } \frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1} \end{array} $$

Step-by-Step Solution

Answer

The formula for the sum is \( \frac{n}{n+1} \).

Step 1: Understand the Problem

We need to find a formula for the sum \[ \sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)} \] by recognizing the pattern in the hint provided.

Step 2: Use the Hint to Simplify Terms

The hint provided states \( \frac{1}{i(i+1)} = \frac{1}{i} - \frac{1}{i+1} \). Use this to rewrite each term in the sum. Thus, each term \( \frac{1}{i(i+1)} \) can be written as \( \frac{1}{i} - \frac{1}{i+1} \).

Step 3: Write the Series Using the Hint

Substitute the identity into the sum:\[\sum_{i=1}^{n} \left( \frac{1}{i} - \frac{1}{i+1} \right)\]Now, list out the first few terms to notice a pattern: \( \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n+1} \right) \).

Step 4: Identify the Telescoping Pattern

Observe that this sum is telescoping, meaning most intermediate terms cancel out. Specifically, each negative term \(-\frac{1}{i+1}\) cancels with part of the previous positive term \(\frac{1}{i+1}\).

Step 5: Simplify the Sum

After cancelling terms in the telescopic sum, only the first positive and the last negative terms remain: \( \frac{1}{1} - \frac{1}{n+1} \).

Step 6: Find the Nice Formula

The simplified formula for the sum is:\[ 1 - \frac{1}{n+1} = \frac{n}{n+1} \]

Step 7: Conclude the Solution

The formula for the given series \( \sum_{i=1}^{n} \frac{1}{i(i+1)} \) is \( \frac{n}{n+1} \).

Key Concepts

Partial Fraction Decomposition

Partial Fraction Decomposition is a technique used to break down complex fractions into simpler ones. It's particularly useful when we're dealing with rational expressions that we want to simplify. Let's think of it like turning a difficult jigsaw puzzle into smaller, more manageable pieces—as the goal is to find out what those pieces are first.
In our problem, the expression \( \frac{1}{i(i+1)} \) can be decomposed using partial fractions. The hint in the exercise suggests writing it as:

• \( \frac{1}{i(i+1)} = \frac{1}{i} - \frac{1}{i+1} \)

This decomposes our fraction into two simpler fractions. Understanding how to make this decomposition is vital because it makes subsequent steps much easier. The result is that we can rewrite every term in the series, which is a crucial step in simplifying the sum.

Series Summation

Series Summation involves finding the sum of a sequence of terms. Sometimes this sequence seems daunting, but certain techniques can make it way simpler. Essentially, series summation is about adding up the many small parts into one whole.
The given series \( \sum_{i=1}^{n} \left( \frac{1}{i(i+1)} \right) \) is simplified using our earlier partial fraction decomposition. After the decomposition, the task becomes adding these individual fractions: \( \sum_{i=1}^{n} \left( \frac{1}{i} - \frac{1}{i+1} \right) \).

• This setup allows us to identify a pattern where elements cancel each other out in what we call a telescoping series.

When written out, notice how terms in the expanded series cancel each other, simplifying the entire sum to just the non-canceled elements.

• Final result: \( 1 - \frac{1}{n+1} \)

This telescoping property significantly reduces otherwise complicated calculations.

Mathematical Induction

Mathematical Induction is a proof technique much like climbing a ladder: you prove it works for the first step, and assuming it holds for one step ensures it works for the next. This concept is invaluable for verifying formulas derived from series summations.
For confirming our derived formula, \( \frac{n}{n+1} \), mathematical induction can be applied as follows:

• **Base Step:** Verify the formula is correct for an initial value, such as \( n = 1 \). If you calculate directly, \( \frac{1}{1 \cdot 2} = \frac{1}{2} \), which corresponds to \( 1 - \frac{1}{2} \).
• **Inductive Step:** Assume the formula holds for some arbitrary \( k = n \), i.e., \( \sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1} \), and prove it holds for \( k = n+1 \).

Using these steps means you're ensuring that, logically, the formula works for all natural numbers, confirming that the original problem's derived results are not just a fluke but robust against deeper scrutiny.