Problem 23 - Calculus Single Variable

Question

A series \(\sum_{n=1}^{\infty} a_{n}\) is given. Calculate the first five partial sums of the series. That is, calculate \(S_{N}=\sum_{n=1}^{N} a_{n}\) for \(N=1,2,3,4,5\). $$ \sum_{n=1}^{\infty}\left(2^{-n}+1\right) $$

Step-by-Step Solution

Answer

The first five partial sums are \( S_1 = \frac{3}{2} \), \( S_2 = \frac{11}{4} \), \( S_3 = \frac{31}{8} \), \( S_4 = \frac{79}{16} \), and \( S_5 = \frac{191}{32} \).

Step 1: Understanding the Series

The given series is \( \sum_{n=1}^{\infty} \left( 2^{-n} + 1 \right) \). This is an infinite series where each term is \( a_n = 2^{-n} + 1 \). To find the partial sums, \( S_N \), we need to calculate the sum of the first \( N \) terms of this series.

Step 2: Calculate the First Term

The first term of the series is when \( n = 1 \). Calculate \( a_1 = 2^{-1} + 1 = \frac{1}{2} + 1 = \frac{3}{2} \). Therefore, the first partial sum \( S_1 = \frac{3}{2} \).

Step 3: Calculate the Second Term

The second term is \( a_2 = 2^{-2} + 1 = \frac{1}{4} + 1 = \frac{5}{4} \). Add to the previous sum to find \( S_2 = S_1 + a_2 = \frac{3}{2} + \frac{5}{4} = \frac{6}{4} + \frac{5}{4} = \frac{11}{4} \).

Step 4: Calculate the Third Term

The third term is \( a_3 = 2^{-3} + 1 = \frac{1}{8} + 1 = \frac{9}{8} \). The third partial sum is \( S_3 = S_2 + a_3 = \frac{11}{4} + \frac{9}{8} = \frac{22}{8} + \frac{9}{8} = \frac{31}{8} \).

Step 5: Calculate the Fourth Term

The fourth term is \( a_4 = 2^{-4} + 1 = \frac{1}{16} + 1 = \frac{17}{16} \). The fourth partial sum is \( S_4 = S_3 + a_4 = \frac{31}{8} + \frac{17}{16} = \frac{62}{16} + \frac{17}{16} = \frac{79}{16} \).

Step 6: Calculate the Fifth Term

The fifth term is \( a_5 = 2^{-5} + 1 = \frac{1}{32} + 1 = \frac{33}{32} \). The fifth partial sum is \( S_5 = S_4 + a_5 = \frac{79}{16} + \frac{33}{32} = \frac{158}{32} + \frac{33}{32} = \frac{191}{32} \).

Key Concepts

Partial Sums

Partial sums are an important concept when working with infinite series. They help us understand how portions of an infinite series add up. You can think of partial sums as a way of summing only the beginning of the series up to a specific point.For the series given in the exercise, each partial sum is expressed as \( S_N = \sum_{n=1}^{N} a_n \), where \( N \) is the number of terms we want to include.In the step-by-step solution, we calculated:

• First Partial Sum \( S_1 \): The sum of just the first term, \( a_1 = 3/2 \).
• Second Partial Sum \( S_2 \): The sum of the first two terms, resulting in \( S_2 = 11/4 \).
• Third Partial Sum \( S_3 \): Adds the third term to get \( 31/8 \).
• Fourth Partial Sum \( S_4 \): Includes the fourth term, making it \( 79/16 \).
• Fifth Partial Sum \( S_5 \): Finally, adds the fifth term totaling \( 191/32 \).

This process helps us approximate how the series grows and gives an idea of its behavior over finite series. It's the foundation for exploring convergence and divergence.

Convergence of Series

The convergence of a series tells us whether a sum of an infinite series approaches a particular number as more terms are added. For infinite series like \( \sum_{n=1}^{\infty} (2^{-n} + 1) \), understanding convergence helps determine if it resolves to a finite value or keeps increasing without bounds.To check convergence, one common method is to evaluate the limit of the partial sums as \( N \to \infty \). If the limit exists and is finite, the series converges. If not, it diverges.For the series in the exercise, each term \( 2^{-n} \) approaches zero as \( n \) increases, but adding 1 to each term affects this process. The added constant ensures the series continually increases since 1 is not vanishing, hinting towards divergence.In general, tests like the Ratio Test, Root Test, or Comparison Test are used for determining convergence.

Arithmetic Sequences

Although the terms in the described series involve powers of two, the concept of arithmetic sequences is still relevant to distinguish patterns in sequences and series.In an arithmetic sequence, the difference between successive terms remains constant. This is not the case for our series, where the function \( a_n = 2^{-n} + 1 \) does not produce such a sequence due to its dependence on exponential decay by powers of two.Despite this, understanding arithmetic sequences can be useful for comparing characteristics of different kinds of series and for constructing series with predictable linear growth. Recognizing these differences enhances comprehension of various infinite sequences and how they sum uniquely.This insight helps identify when a simple pattern is disturbed by exponential or other non-linear changes, which is crucial in distinguishing between arithmetic sequences and other types like geometric or those, like ours, involving additional constants.