Problem 91 - Calculus: Early Transcendentals

Question

Consider the series \( \sum_{n = 1}^{\infty} n/(n + 1)!. \) (a) Find the partial sums \( s_1, s_2, s_3, \) and \( s_4. \) Do you recognize the denominators? Use the pattern to guess a formula for \( s_n. \) (b) Use mathematical induction to prove your guess. (c) Show that the given infinite series is convergent, and find its sum.

Step-by-Step Solution

Answer

The series converges to 1.

Step 1: Find Partial Sum s_1

To find the partial sum \( s_1 \), evaluate the sum when \( n=1 \):\[s_1 = \frac{1}{2!} = \frac{1}{2}.\]

Step 2: Find Partial Sum s_2

Now, evaluate the partial sum \( s_2 \) by summing the first two terms:\[s_2 = \frac{1}{2!} + \frac{2}{3!} = \frac{1}{2} + \frac{2}{6} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}.\]

Step 3: Find Partial Sum s_3

Evaluate \( s_3 \) by including the third term:\[s_3 = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} = \frac{5}{6} + \frac{3}{24} = \frac{5}{6} + \frac{1}{8} = \frac{20}{24} + \frac{3}{24} = \frac{23}{24}.\]

Step 4: Find Partial Sum s_4

Evaluate \( s_4 \) similarly:\[s_4 = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \frac{4}{5!} = \frac{23}{24} + \frac{4}{120} = \frac{23}{24} + \frac{1}{30} = \frac{575}{720} + \frac{24}{720} = \frac{599}{720} = \frac{120}{144} = \frac{5}{6}.\]

Step 5: Observe Denominator Pattern

Noticing the pattern, the denominators observed for \( s_1, s_2, s_3, \) and \( s_4 \) are related to factorial numbers. Assuming a simplification, we guess the general formula for \( s_n \) could be represented as a fraction where the denominator is factorial of one more than the number of terms.

Step 6: Formulate General Formula for s_n

The pattern suggests a formula:\[s_n = 1 - \frac{1}{(n+1)!}.\]

Step 7: Inductive Base Case

Verify the base case \( n=1 \):\[s_1 = 1 - \frac{1}{2!} = \frac{1}{2},\] which holds true as calculated in Step 1.

Step 8: Inductive Hypothesis

Assume the formula holds for n=k:\[s_k = 1 - \frac{1}{(k+1)!}.\]

Step 9: Inductive Step

Prove it for n=k+1 using the hypothesis:Add \( \frac{k+1}{(k+2)!} \) to both sides of \( s_k = 1 - \frac{1}{(k+1)!} \):\[s_{k+1} = 1 - \frac{1}{(k+1)!} + \frac{k+1}{(k+2)!}.\] This simplifies to:\[s_{k+1} = 1 - \frac{1}{(k+2)!}.\] Thus, the inductive step holds.

Step 10: Convergence and Sum of the Series

Since \( s_n = 1 - \frac{1}{(n+1)!} \) converges to 1 as \( n \to \infty \), the infinite series converges. The sum is thus:\[\sum_{n=1}^{\infty} \frac{n}{(n+1)!} = 1.\]

Key Concepts

Partial Sums

Exploring an infinite series often begins by examining its partial sums. A partial sum is the sum of the first few terms of a series. For example, given the series \( \sum_{n = 1}^{\infty} \frac{n}{(n+1)!} \), the partial sums help us glimpse into the behavior of the series.

To find a partial sum, we calculate the sum up to a certain number of terms. Starting with \( s_1 = \frac{1}{2!} \), we evaluate \( s_2 \), which is \( \frac{1}{2!} + \frac{2}{3!} \), and so forth, until we find \( s_3 \) as \( \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} \), and continue in this fashion. By calculating these values, we identify patterns or general formulas, such as the pattern observed in the original solution, where the result simplifies to a form involving factorials like \( \frac{1}{(n+1)!}.\)

Understanding partial sums is crucial because they give insights into how the series behaves as it progresses towards more terms. By writing each term incrementally, it's possible to discover potential simplifications. And hence, these partial sums can help in predicting if the series will converge and what the sum might be.

Mathematical Induction

Mathematical induction is a powerful tool for proving statements about natural numbers. In the context of our series, once we have guessed a formula for the partial sums, we can use induction to prove that this formula works for all terms.

The process of mathematical induction generally involves three steps:

• **Base Case**: Verify the formula for the smallest value, often \( n = 1 \).
• **Inductive Hypothesis**: Assume the formula is true for \( n = k \).
• **Inductive Step**: Prove it must also be true for \( n = k+1 \).

By successfully completing these steps, as shown with the formula \( s_n = 1 - \frac{1}{(n+1)!} \), we ensure that our guess applies to every term in the series. For example, starting with the base case \( s_1 = 1 - \frac{1}{2!} \) and assuming the formula works for \( s_k = 1 - \frac{1}{(k+1)!} \), we add \( \frac{k+1}{(k+2)!} \) to reach \( s_{k+1} \) which smoothly simplifies to \( 1 - \frac{1}{(k+2)!} \), thus completing the proof.

This step not only solidifies our findings but also reinforces the reliability of the pattern we identified within the partial sums.

Series Convergence

Series convergence is a fundamental concept that tells us whether the sum of an infinite series approaches a specific value as more terms are added. In simpler terms, it's the idea of whether or not the series "settles down".

For our series \( \sum_{n=1}^{\infty} \frac{n}{(n+1)!} \), convergence can be checked using the formula for partial sums derived earlier. We found \( s_n = 1 - \frac{1}{(n+1)!} \), showcasing that as \( n \) approaches infinity, \( \frac{1}{(n+1)!} \) becomes increasingly negligible, leading \( s_n \) closer to 1.

A convergent series means you can actually assign a definitive limit to its sum. It's akin to saying that despite being "infinite", the numerical value stabilizes at a particular number. For this series, that number, or the sum, is 1. Therefore, understanding series convergence helps us not only predict the behavior of the series but also enables us to ascertain the converged value or sum.

In practical applications, this informs us about how systems described by such series behave over time or in processes that iterate indefinitely.