Problem 1 - Calculus: One and Several Variables

Question

Find the sum of the series. $$\sum_{k=1}^{\infty} \frac{1}{2 k(k+1)}$$

Step-by-Step Solution

Answer

The sum of the given infinite series is equal to \(\frac{1}{2}\).

Step 1: Identify the type of series

We are dealing with an infinite series of the following form: $$\sum_{k=1}^{\infty} \frac{1}{2 k(k+1)} = \frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k (k+1)}$$

Step 2: Perform partial fraction decomposition

We notice that the term inside the summation is a fraction, and we can use partial fraction decomposition to simplify it: $$\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}$$ We want to find constants A and B. To do so, we get rid of the denominators first by multiplying with the common denominator \(k(k+1)\): $$1 = A(k+1) + Bk$$ By solving this equation for the constants, we arrive at the following step:

Step 3: Solve for constants A and B

To solve for A, set \(k=0\): $$1 = A(0+1) => A=1$$ To solve for B, substitute the value of A back into the equation and set \(k=-1\): $$1 = (1)(-1+1) + B(-1) => B=-1$$ Now we can rewrite the original series as a sum of simpler expressions: $$\frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k (k+1)} = \frac{1}{2}\sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right)$$

Step 4: Simplify the series using telescoping

This is a telescoping series, meaning that most of its terms cancel out each other when written in expanded form: $$\frac{1}{2}\left[\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dotsb\right]$$ In the parentheses, the second term of each pair cancels the first term of the next pair, and so on. This cancellation leaves only the first term of the first pair as a significant term.

Step 5: Find the sum of the series

After the cancellations, the sum of the series can be found as follows: $$\frac{1}{2}\left(\frac{1}{1}\right) = \boxed{\frac{1}{2}}$$ Therefore, the sum of the given infinite series is equal to \(\frac{1}{2}\).

Key Concepts

Partial Fraction Decomposition

When you encounter a complex rational function within an infinite series, partial fraction decomposition simplifies the problem. For our series \(\sum_{k=1}^{\infty} \frac{1}{2k(k+1)}\), we begin by expressing each term as a simpler fraction. The idea is to break down \(\frac{1}{k(k+1)}\) into the sum of fractions like \(\frac{A}{k} + \frac{B}{k+1}\). This technique transforms the problem into finding the constants \(A\) and \(B\).
To find these constants, you eliminate the denominators by multiplying throughout by the common denominator, which is \(k(k+1)\). This gives us the equation:

• \(1 = A(k+1) + Bk\)

By clever substitution choices—such as \(k=0\) and \(k=-1\)—you can solve for \(A\) and \(B\), yielding \(A=1\) and \(B=-1\). Hence, the term \(\frac{1}{k(k+1)}\) is equivalent to \(\frac{1}{k} - \frac{1}{k+1}\). This simplification is crucial for the next step.

Telescoping Series

A telescoping series cleverly cancels out terms to simplify sums. After applying partial fraction decomposition, our series becomes:

• \(\frac{1}{2}\sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right)\)

When you expand such series, every negative term from one fraction matches with the positive term of the next. This means most terms "telescope" or cancel out:

• \(\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots\)

In this pattern, after cancellation, you're left with just the first term from the initial fraction sequences. This property makes telescoping series particularly useful for evaluating the sum of infinite series.

Series Convergence

The culmination of this exercise is understanding series convergence. Many series might not converge, but when they do, they approach a specific value. Our transformed telescoping series, after simplification, boils down to:

• \(\frac{1}{2} \times \left( \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right) \right)\)

As \(n\) approaches infinity, \(\frac{1}{n+1}\) shrinks to zero, leaving us with just \(\frac{1}{2} \times 1\), which is \(\frac{1}{2}\).
This example highlights that the series converges to \(\frac{1}{2}\). Understanding convergence involves checking whether terms within the series shrink as the index grows indefinitely, which determines if they approach a finite sum or diverge. For the telescoped expression, the series converges nicely, illustrating how mathematical manipulations can lead to elegant solutions.