Question

A series is given. (a) Find a formula for \(S_{n},\) the \(n^{\text {th }}\) partial sum of the series. (b) Determine whether the series converges or diverges. If it converges, state what it converges to. $$\sum_{n=1}^{\infty} e^{-n}$$

Short answer

(a) Partial sum: \( S_n = \frac{1-(\frac{1}{e})^n}{e-1} \); (b) Series converges to \( \frac{1}{e-1} \).

Step-by-step solution

Writing the Series

The series given is the sum of exponential terms: \( \sum_{n=1}^{\infty} e^{-n} \). This series starts at \( n = 1 \) and continues indefinitely.

Understanding the Series Type

Each term in the series is of the form \( e^{-n} \), which can also be written as \( \left( \frac{1}{e} \right)^{n} \). This is a geometric series with the first term \( a = e^{-1} \) and the common ratio \( r = \frac{1}{e} \).

Finding the Partial Sum Formula

For a geometric series, the \(n^{\text{th}}\) partial sum \( S_n \) is given by the formula: \[ S_n = a \frac{1-r^n}{1-r} \] where \( a = e^{-1} \) and \( r = \frac{1}{e} \). Substituting, we have: \[ S_n = e^{-1} \frac{1-(\frac{1}{e})^n}{1-\frac{1}{e}} \].

Simplifying the Partial Sum Formula

In the expression \( S_n = e^{-1} \frac{1-(\frac{1}{e})^n}{1-\frac{1}{e}} \), simplify \( 1 - \frac{1}{e} \) to \( \frac{e-1}{e} \). Substituting this back in, we have: \[ S_n = e^{-1} \frac{1-(\frac{1}{e})^n}{\frac{e-1}{e}} = \frac{1-(\frac{1}{e})^n}{e-1} \].

Determining Convergence or Divergence

A geometric series converges if the absolute value of the common ratio \( r \) is less than 1. Here, \( r = \frac{1}{e} \) and since \( e > 1 \), \( \frac{1}{e} < 1 \), so the series converges.

Finding the Limit as \(n \to \infty\)

Since the series converges, we calculate \( \lim_{n \to \infty} S_n \): \[ \lim_{n \to \infty} \frac{1-(\frac{1}{e})^n}{e-1} \]. As \( n \to \infty \), \( (\frac{1}{e})^n \to 0 \). Therefore, \[ \lim_{n \to \infty} S_n = \frac{1-0}{e-1} = \frac{1}{e-1} \].

Key concepts

Partial Sum Formula

In the world of series, a geometric series holds a special place due to its simplicity and elegance. The partial sum formula of a geometric series helps you find the sum of a fixed number of terms.
For a geometric series, where each term is of the form \( ar^n \), the partial sum formula is:

• \( S_n = a \frac{1-r^n}{1-r} \)

Here, \( a \) is the first term and \( r \) is the common ratio. This formula is useful because it helps to quickly sum up the series up to a certain number \( n \) of terms.
In our exercise, the series \( \sum_{n=1}^{\infty} e^{-n} \) translates to a geometric series with:

• \( a = e^{-1} \)
• \( r = \frac{1}{e} \)

Substituting these into the partial sum formula gives:\[S_n = e^{-1} \frac{1-(\frac{1}{e})^n}{1-\frac{1}{e}}\]
After simplification, it further reduces to:
\[S_n = \frac{1-(\frac{1}{e})^n}{e-1}\]Understanding this formula is crucial for analyzing how the series behaves as you sum more terms.

Convergence and Divergence

Convergence and divergence play a key role in analyzing infinite series. A series converges if adding up all its terms leads to a finite number. Conversely, a series diverges if the sum doesn't settle into a number.
To determine convergence for a geometric series, consider the absolute value of the common ratio \( r \):

• If \( |r| < 1 \), the series converges.
• If \( |r| \geq 1 \), the series diverges.

For our series \( \sum_{n=1}^{\infty} \left( \frac{1}{e} \right)^n \), the common ratio \( r = \frac{1}{e} \). Since \( e \approx 2.718 > 1 \), \( \frac{1}{e} < 1 \). Hence, the series converges.
When a series converges, it means that as you add more terms, the sum approaches a finite limit. In this exercise, the series converges to \( \frac{1}{e-1} \) as calculated through the partial sum formula as \( n \to \infty \). This is essential in many mathematical contexts, ensuring calculations involving such series are meaningful and valid.

Exponential Series

Exponential series involve terms where the variable is in the exponent, like in our example \( \sum_{n=1}^{\infty} e^{-n} \). Such series are common in mathematics and help represent complex situations in a simpler form.
The terms of our given series \( e^{-n} \) can be rewritten as \( \left( \frac{1}{e} \right)^n \), showing it's a geometric series. The exponential decay nature comes from the negative exponent, indicating the size of terms decreases rapidly as \( n \) increases.
Exponential series can model real-world phenomena like:

• Radioactive decay
• Population growth or decline
• Interest compounding in finance

Understanding exponential series helps in creating models that predict or describe dynamic systems in science and engineering.
In this exercise, recognizing \( e^{-n} \) as part of an exponential series allowed us to use the knowledge of geometric series for convergence. This representation simplifies finding the sum and determining convergence efficiently.