Question

Find a formula for \(\sum_{n=1}^{\infty} \frac{1}{n(n+N)}\) where \(N\) is a positive integer.

Short answer

\( \sum_{n=1}^{\infty} \frac{1}{n(n+N)} = \frac{1}{N} H_N \).

Step-by-step solution

Break Down the Series

The series given is \( \sum_{n=1}^{\infty} \frac{1}{n(n+N)} \). First, notice that the expression \( \frac{1}{n(n+N)} \) can be simplified using partial fraction decomposition.

Apply Partial Fraction Decomposition

We want to express \( \frac{1}{n(n+N)} \) as a sum of simpler fractions: \( \frac{1}{n(n+N)} = \frac{A}{n} + \frac{B}{n+N} \). Solving, we find \( A = \frac{1}{N} \) and \( B = -\frac{1}{N} \). Thus, \( \frac{1}{n(n+N)} = \frac{1}{N} \left( \frac{1}{n} - \frac{1}{n+N} \right) \).

Simplify the Series

Substituting the partial fraction decomposition into the series, we have: \[ \sum_{n=1}^{\infty} \frac{1}{n(n+N)} = \sum_{n=1}^{\infty} \frac{1}{N} \left( \frac{1}{n} - \frac{1}{n+N} \right). \]

Apply Telescoping Series Property

The series is now a telescoping series. Most terms cancel out: \( \frac{1}{1} - \frac{1}{1+N} \), \( \frac{1}{2} - \frac{1}{2+N} \), and so on. What remains are the first N terms of the first part of each pair: \( \sum_{n=1}^{N} \frac{1}{n} \).

Write the Final Formula

The limit as n approaches infinity of all terms \( \frac{1}{n+N} \) becomes zero. Therefore, the sum of the series is \( \frac{1}{N} \sum_{n=1}^{N} \frac{1}{n} \). Therefore, the formula for the series is: \[ \sum_{n=1}^{\infty} \frac{1}{n(n+N)} = \frac{1}{N} H_N, \] where \( H_N \) is the N-th harmonic number defined as \( H_N = \sum_{n=1}^{N} \frac{1}{n} \).

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in calculus and algebra to break down complex rational expressions into simpler ones. This is especially useful when dealing with infinite series. By decomposing fractions, we can simplify the summation and make it easier to analyze and solve the series.

In the given problem, we start with the expression \( \frac{1}{n(n+N)} \). This can be tricky due to its structure. The idea is to express it as a sum of simpler fractions of the form \( \frac{A}{n} + \frac{B}{n+N} \).

• First, we equate \( \frac{1}{n(n+N)} \) to \( \frac{A}{n} + \frac{B}{n+N} \).
• Then, multiply through by \( n(n+N) \) to clear the denominators, resulting in the equation \( 1 = A(n+N) + Bn \).
• Solving this equation for \( A \) and \( B \), we find that \( A = \frac{1}{N} \) and \( B = -\frac{1}{N} \).

Thus, the given expression becomes \( \frac{1}{n(n+N)} = \frac{1}{N} \left( \frac{1}{n} - \frac{1}{n+N} \right) \). This decomposition sets the stage for applying the telescoping series principle.

Telescoping Series

Telescoping series are a type of infinite series where successive terms cancel each other out. This property significantly simplifies finding the sum of the series.

After applying partial fraction decomposition to our series \( \sum_{n=1}^{\infty} \frac{1}{n(n+N)} \), we rewrite it as \( \sum_{n=1}^{\infty} \frac{1}{N} \left( \frac{1}{n} - \frac{1}{n+N} \right) \).

When expanded, the series looks like this:

• \( \frac{1}{1} - \frac{1}{1+N} \)
• \( \frac{1}{2} - \frac{1}{2+N} \)
• ...and so on.

Notice how every \( \frac{1}{k+N} \) term cancels with the \( -\frac{1}{k+N} \) term in the next pair of terms. This characteristic of alternating sign differences makes most terms cancel, leaving only the initial few from the decomposition.

As a result of this cancellation, the sum of the series simplifies to \( \frac{1}{N} \sum_{n=1}^{N} \frac{1}{n} \). This is where the concept of harmonic numbers comes into play.

Harmonic Numbers

Harmonic numbers are a sequence of numbers that are essential in various areas of mathematics, including number theory and analysis. Defined by the formula \( H_N = \sum_{n=1}^{N} \frac{1}{n} \), harmonic numbers represent the sum of the reciprocals of the first \( N \) natural numbers.

In the context of our series \( \sum_{n=1}^{\infty} \frac{1}{n(n+N)} \), after applying partial fraction decomposition and recognizing the telescoping nature, we arrive at a formula involving harmonic numbers. The series can be expressed ultimately as \( \frac{1}{N} H_N \).

This result leverages the simple form of harmonic numbers' sum, effectively capturing the behavior of the series in a formula that is manageable and elegant. Because harmonic numbers grow slowly compared to other sequences, they provide insight into the convergence and behavior of the series.