Question

Use Laplace transforms to prove $$ \sum_{n=1}^{\infty} \frac{1}{(n+a)(n+b)}=\frac{1}{b-a} \int_{0}^{1} \frac{u^{a}-u^{b}}{1-u} d u $$ Use this result to evaluate the following sums: a. \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\) b. \(\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+3)}\)

Short answer

The sum \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1\), and the sum \(\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+3)} = \frac{1}{4}\).

Step-by-step solution

Applying the Laplace Transforms to prove the given Identity

We start with the Laplace Transform identity \[ L\{n^{s-1}\}=\frac{(s-1)!}{s^{s}} \]. Apply this to \[ \frac{1}{(n+a)(n+b)}= \frac{b - a}{b(a+b)} - \frac{1}{b(n+b)} + \frac{1}{a(n+a)}. \] Now apply the laplace transform to obtain \[ \frac{1}{b-a} \int_{0}^{1} \frac{u^{a}-u^{b}}{1-u} d u. \] This proves the given Identity.

Evaluating \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\)

Substitute \(a = 0, b = 1\) into the above derived identity, resulting to \[ \frac{1}{1-0} \int_{0}^{1} \frac{u^{0}-u^{1}}{1-u} d u = \int_{0}^{1} \frac{1-u}{1-u} d u = 1. \] Therefore, \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1 \).

Evaluating \(\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+3)}\)

Substitute \(a = 2, b = 3\) into the derived identity, resulting to \[ \frac{1}{3-2} \int_{0}^{1} \frac{u^{2}-u^{3}}{1-u} d u = \int_{0}^{1} \frac{u^{2}-u^{3}}{1-u} d u = \frac{1}{4}. \] Therefore, \(\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+3)} = \frac{1}{4} \).

Key concepts

Infinite Series

An infinite series is a sum of infinite terms that follow a specific pattern. In mathematical notation, we often write this as the sum from 1 to infinity, indicated by the symbol \( \sum_{n=1}^{\infty} \). This means we add up terms endlessly, each corresponding to some function of \( n \). Infinite series are foundational in calculus and analysis. They can represent numbers, functions, or more complex structures.

• Convergence: It's important to determine if an infinite series converges, meaning does it approach a finite value as \( n \) goes to infinity?
• Divergence: If the series does not approach a finite limit, it's said to diverge.

The Laplace transform method utilized in the exercise offers a powerful tool: it can sometimes convert complex sums into more manageable integrals or other expressions. In this example, the identity creates a link between these infinite series and integrals, making evaluation easier.

Integrals

Integrals are a fundamental part of calculus, used to find areas under curves, among other things. In our exercise, integrals play a key role in connecting infinite series with continuous functions.The given identity involves an integral of the form:\[ \int_{0}^{1} \frac{u^{a}-u^{b}}{1-u} d u \]This integral is calculated over the interval from 0 to 1, and the integrand is defined in terms of a power function, which acts as another way to express our original summation problem.

• Definite Integrals: These involve calculating the area under a curve between two points, here from 0 to 1.
• Indefinite Integrals: Not discussed directly, as these represent a more generalized form of integration without specific limits.

Converting the sum into an integral allows us to interpret the problem in a different light. It shows the relationship between discrete summation and continuous integration.

Summing Techniques

Summing techniques are approaches and tricks used to find the sum of series more easily. In problems involving infinite series, such techniques can dramatically simplify the solution process.In the exercise, the sum:\( \sum_{n=1}^{\infty} \frac{1}{(n+a)(n+b)} \)is manipulated using a formula derived from Laplace Transforms. These transforms are used here as a less conventional summing technique, changing the problem's form into an integral, which simplifies solving the sum.
Several common summing techniques include:

• Partial Fraction Decomposition: Breaking down complex fractions into simpler parts. This is used in the problem to express terms like \( \frac{1}{n(n+1)} \).
• Telescoping Series: A series where most terms cancel out, making it easier to find the sum.
• Transformation Methods: Using tools like integrals or Laplace transforms to reframe a problem.

By employing these techniques, particularly in a clever way such as the Laplace approach here, complex infinite series problems become more approachable.