Problem 41 - Calculus: Early Transcendentals

Question

Determine whether the series is convergent or divergent. If it is convergent, find its sum. \( \displaystyle \sum_{n = 1}^{\infty} \left( \frac {1}{e^n} + \frac {1}{n(n + 1)} \right) \)

Step-by-Step Solution

Answer

The series converges with a sum of \( \frac{1}{e-1} + 1 \).

Step 1: Split the Series

The series can be written as the sum of two separate series: \( \sum_{n = 1}^{\infty} \frac{1}{e^n} + \sum_{n = 1}^{\infty} \frac{1}{n(n+1)} \). We will analyze these two series separately to determine their convergence.

Step 2: Analyze the First Series

Consider the first series \( \sum_{n=1}^{\infty} \frac{1}{e^n} \). This is a geometric series with the first term \( a = \frac{1}{e} \) and common ratio \( r = \frac{1}{e} \), where \( |r| < 1 \). Thus, this series is convergent.

Step 3: Calculate the Sum of the First Series

The sum of an infinite geometric series is given by \( \frac{a}{1-r} \). Here, \( a = \frac{1}{e} \) and \( r = \frac{1}{e} \), yielding the sum:\[ \frac{\frac{1}{e}}{1 - \frac{1}{e}} = \frac{1}{e-1} \].

Step 4: Analyze the Second Series

Consider the second series \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \). This can be decomposed using partial fraction decomposition:\[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \].This is a telescoping series.

Step 5: Calculate the Sum of the Second Series

The telescoping nature of the series \( \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) \) results in cancellation of terms. The series converges to:\[ 1 \].

Step 6: Sum the Results

To find the sum of the original series, combine the results of the two parts: \[ \frac{1}{e-1} + 1 \].Thus, the series converges and its sum is \( \frac{1}{e-1} + 1 \).

Key Concepts

Geometric Series

When we talk about geometric series, we're referring to a series where each term is a constant multiple (called the common ratio) of the previous term. Understanding geometric series is crucial because of their simplicity and wide applications. In the series \( \sum_{n=1}^{\infty} \frac{1}{e^n} \), each term can be expressed as \( \frac{1}{e} \cdot (\frac{1}{e})^{n-1} \), showing it's a geometric series.
The first term \( a \) is \( \frac{1}{e} \), and the common ratio \( r \) is also \( \frac{1}{e} \). A geometric series converges if the absolute value of the common ratio is less than 1, \( |r| < 1 \). Here, since \( \frac{1}{e} < 1 \), series convergence is guaranteed.
To find the sum of an infinite geometric series with a convergent ratio, use the formula \( \frac{a}{1-r} \). For this series:

• The common ratio \( r \) is \( \frac{1}{e} \).
• The sum is then \( \frac{\frac{1}{e}}{1 - \frac{1}{e}} \).
• This simplifies to \( \frac{1}{e-1} \).
Hence, the sum is \( \frac{1}{e-1} \)."

Telescoping Series

A telescoping series is a series where most terms cancel out when the series is expanded. This can make them much simpler to work with.
Consider the series \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \). After partial fraction decomposition, it becomes \( \sum_{n=1}^{\infty} (\frac{1}{n} - \frac{1}{n+1}) \).
This series is telescoping because:

• Consecutive terms cancel out, leaving only a few terms behind.
• As \( n \) increases, most \( \frac{1}{n} \) terms cancel with \( \frac{1}{n+1} \) from the next step.

The beauty lies in its simplicity. With telescoping series like this, the sum is simply determined by the uncancelled terms.
For this series, after cancellation, you’re left with the first term \( 1 \). So, the summed result is \( 1 \)."

Partial Fraction Decomposition

Partial fraction decomposition is a technique for breaking down a fraction into simpler pieces, making complex series easier to handle.
In the given series \( \frac{1}{n(n+1)} \), notice the expression looks complex at first glance. The partial fraction decomposition allows us to write it as a difference between two fractions:

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

This technique simplifies terms and is especially handy in creating telescoping series, where the decomposition lets individual terms cancel each other out.
Mastering partial fractions helps you break down and simplify intricate expressions into manageable components, revealing broader patterns or simplifications inherent in series."

Infinite Series Sum

Understanding infinite series sum is essential in calculus and higher mathematics. An infinite series is essentially the sum of an infinite sequence of numbers. It can either converge to a finite value or diverge to infinity.
In our example, we looked at the sum of two infinite series separately:

• A convergent geometric series \( \sum_{n=1}^{\infty} \frac{1}{e^n} \) which summed to \( \frac{1}{e-1} \).
• A convergent telescoping series \( \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) \) which simplified to \( 1 \).

By combining these results, the overall series' sum is \( \frac{1}{e-1} + 1 \).
The process shows us that by identifying components of an infinite series, utilizing convergence tests, and appropriate simplifying techniques, we can find finite sums for apparently complex infinite series."