Question

In the following exercises, compute the general term an of the series with the given partial sum Sn. If the sequence of partial sums converges, find its limit S. $$ S_{n}=2-(n+2) / 2^{n}, n \geq 1 $$

Short answer

The general term \( a_n = \frac{n}{2^n} \) and the series converges to 2.

Step-by-step solution

Understanding Partial Sums

We are given the partial sum of a series as \( S_n = 2 - \frac{n+2}{2^n} \). The task is to identify the general term \( a_n \) of the series and determine if the sequence of partial sums \( S_n \) converges.

Determine General Term of the Series

The general term \( a_n \) of a series is found by calculating the difference between consecutive terms of the partial sums. Therefore, \( a_n = S_n - S_{n-1} \).

Calculate \( S_{n-1} \)

Substitute \( n-1 \) into the expression for \( S_n \): \( S_{n-1} = 2 - \frac{(n-1) + 2}{2^{n-1}} = 2 - \frac{n+1}{2^{n-1}} \).

Calculate Difference to Find \( a_n \)

Now find the general term \( a_n \) by calculating \( S_n - S_{n-1} \):\( a_n = \left(2 - \frac{n+2}{2^n}\right) - \left(2 - \frac{n+1}{2^{n-1}}\right) \).

Simplify \( a_n \) Expression

Simplify the expression:\( a_n = -\frac{n+2}{2^n} + \frac{n+1}{2^{n-1}} = -\frac{n+2}{2^n} + \frac{2(n+1)}{2^n} \), which reduces to \( a_n = \frac{n}{2^n} \).

Check for Convergence of Series

To check if the partial sums \( S_n \) converge, take the limit as \( n \) approaches infinity:\( \lim_{{n \to \infty}} S_n = \lim_{{n \to \infty}} \left(2 - \frac{n+2}{2^n}\right) \).

Calculate the Limit of \( S_n \) as \( n \to \infty \)

Evaluate \( \lim_{{n \to \infty}} -\frac{n+2}{2^n} \). The exponential term \( 2^n \) grows much faster than any linear term \( n \), hence it approaches zero. Therefore, \( \lim_{{n \to \infty}} S_n = 2 - 0 = 2 \).

Key concepts

Series Convergence

Series convergence is a crucial concept in the realm of calculus and analysis. It helps us understand whether a series sums up to a finite value as more terms are added. In the context of arithmetic, when we say a series converges, we're basically saying that the sum of its terms approaches a specific number, called the limit. Series convergence comes in handy in calculations and predictions, especially when dealing with infinite series. To determine convergence, we often look at the sequence of partial sums of the series. If the partial sums tend toward a finite value as the number of terms increases, the series is said to be convergent. An important tool for studying convergence is the limit of the sequence of partial sums. If the limit exists and is finite, the series converges to that limit. The notion of convergence plays a pivotal role in many areas of mathematics, including the development of functions, Fourier series, and even in complex analysis.

General Term of Series

The general term of a series, often denoted as \( a_n \), represents the formula for any given term in the series as a function of \( n \). Finding this formula is key to understanding the behavior of the series. To determine the general term, we explore the given partial sums, which are the cumulative sums up to that point in the sequence. The general term can be found by examining the difference between consecutive partial sums: \( a_n = S_n - S_{n-1} \), where \( S_n \) is the nth partial sum. For the series given in the exercise, after computing \( S_n - S_{n-1} \), we identify the general term as \( a_n = \frac{n}{2^n} \). This formula provides the specific value of the nth term of the series and gives insight into how each term contributes to the overall sum. Understanding the general term is essential for calculating the sum of the series and for determining whether the series converges.

Limit Calculation

The concept of limit calculation is foundational in determining series convergence. It's all about understanding what value the sequence of partial sums (or other functions) approaches as the index grows infinitely large. Limit calculation applies here where we have to find \( \lim_{{n \to \infty}} S_n \). With the exercise's series, the partial sum is given as \( S_n = 2 - \frac{n+2}{2^n} \). As \( n \to \infty \), the term \( \frac{n+2}{2^n} \) gets closer to 0 since the denominator \( 2^n \) grows exponentially faster than the numerator. The crucial insight is that as the variable grows larger, the components that decrease faster tend to drop out in the limit context. Hence, \( \lim_{{n \to \infty}} S_n = 2 \). This calculation assures us that the original series converges to the limit \( 2 \). Knowing how to calculate limits is a pivotal skill in calculus, essential for mastering series analysis, integration, and solving differential equations.