/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Consider the series $$ S_{n}... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the series $$ S_{n}=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\ldots+\frac{n}{2^{n}} $$ Show that $$ \frac{1}{2} S_{n}=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\ldots+\frac{n}{2^{n+1}} $$ and hence that $$ S_{n}-\frac{1}{2} S_{m}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots+\frac{1}{2^{n}}+\frac{n}{2^{n+1}} $$ Hence sum the series.

Short Answer

Expert verified
The series sums up to 2 as \( n \to \infty \).

Step by step solution

01

Write Down Expressions for Original Series and Half Series

The original series is given by \( S_{n} = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \ldots + \frac{n}{2^{n}} \). We want to find \( \frac{1}{2} S_{n} \), which implies multiplying each term of the series by \( \frac{1}{2} \). So, \( \frac{1}{2} S_{n} = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \ldots + \frac{n}{2^{n+1}} \).
02

Subtract Half Series from Original Series

Subtract \( \frac{1}{2} S_{n} \) from \( S_{n} \) to simplify the terms.Starting with the terms:- \( S_{n} = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \ldots + \frac{n}{2^{n}} \)- \( \frac{1}{2} S_{n} = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \ldots + \frac{n}{2^{n+1}} \)Performing the subtraction yields:\[S_{n} - \frac{1}{2} S_{n} = \left( \frac{1}{2} - 0 \right) + \left( \frac{2}{4} - \frac{1}{4} \right) + \left( \frac{3}{8} - \frac{2}{8} \right) + \ldots + \left( \frac{n}{2^{n}} - \frac{n}{2^{n+1}} \right)\]This simplifies to:\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^{n}} + \frac{n}{2^{n+1}}\]
03

Recognize and Simplify the Resultant Series

The resultant expression from the previous step is a geometric series plus an additional term:\[S_{n} - \frac{1}{2} S_{n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^{n}} + \frac{n}{2^{n+1}}\]The geometric series \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^{n}} \) has a common ratio of \( \frac{1}{2} \) and first term \( \frac{1}{2} \). The sum of this geometric series is given by:\[\frac{1}{2} \left( \frac{1 - \left(\frac{1}{2}\right)^{n}}{1 - \frac{1}{2}} \right) = 1 - \frac{1}{2^{n}}\]Therefore, \[S_{n} - \frac{1}{2} S_{n} = 1 - \frac{1}{2^{n}} + \frac{n}{2^{n+1}}\]
04

Solve for Final Series Total

Based on Step 3, we have:\[\frac{1}{2} S_{n} = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \ldots + \frac{n}{2^{n+1}}\]Multiply both sides of this equation by 2 to find \( S_{n} \):\[S_{n} = 2 \left( 1 - \frac{1}{2^{n}} + \frac{n}{2^{n+1}} \right)\]
05

Conclude the Series Sum

As \( n \rightarrow \infty \), the term \( \frac{1}{2^{n}}\) approaches zero, simplifying:\[S_{n} \approx 2\]Thus, the series effectively sums up to 2 as \( n \) becomes very large.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series Manipulation
In mathematics, series manipulation involves altering a series without changing its overall value. This is done to simplify calculations or to reveal underlying patterns. In the exercise above, manipulation helps in reaching a form where it's easier to work with the series. It starts by taking the given series and creating a half series, where each term is multiplied by 1/2. This results in:
  • Original series: \( S_{n} = \frac{1}{2} + \frac{2}{4} + \ldots + \frac{n}{2^{n}} \)
  • Half series: \( \frac{1}{2}S_{n} = \frac{1}{4} + \frac{2}{8} + \ldots + \frac{n}{2^{n+1}} \)

Subtracting the half series from the original one leads to further simplification, forming another recognizable series. This manipulation not only simplifies the series but also helps unveil a geometric component, essential for finding the sum of the series.
Geometric Progression
A geometric progression, or a geometric series, is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In our exercise, the series extracted from manipulation exhibits this behavior. We have:
  • First term (\(a\)): \( \frac{1}{2} \)
  • Common ratio (\(r\)): \( \frac{1}{2} \)

These characteristics result in the series:
  • \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^{n}} \)
This is a classic geometric progression. The sum of a finite geometric series is given by the formula:\[S = a \frac{1 - r^n}{1 - r}\]Applying this to our series: \[S = \frac{1}{2} \left( \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} \right)\]This formula helps us find the sum, emphasizing the power of geometric progression in simplifying complex series.
Infinite Series
An infinite series is a series that continues indefinitely. In this context, we're interested in what happens as the number of terms, \(n\), approaches infinity. For the series to converge (reach a sum), its terms must approach zero. In the given problem, as \(n\) becomes very large, the term \(\frac{1}{2^n}\) nears zero.
This behavior allows us to determine that, even though the series \(S_n\) starts as a finite geometric progression, it extends to an infinite series as \(n\) approaches infinity. In this scenario, the sum of the series can be approximated by:
  • \(S_n \approx 2\)

The concept of infinite series is crucial in calculus and mathematical analysis, offering a way to sum sequences that do not have a definite end. Understanding this allows students to evaluate limits and recognize convergent behavior in series, as the sum effectively "stabilizes" at a particular value, despite the infinite nature of the series.

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