**Analyzing Convergence of a Sequence****Problem Statement:**Does the sequence converge? If so, what is its limit? If not, why not?\[ \left\{ \frac{ne^{-n}}{n + e^{-n}} \right\}_{n=1}^{\infty} \]**Explanation:**We are given a sequence defined by the formula \( \frac{ne^{-n}}{n + e^{-n}} \) and we need to determine whether this sequence converges as \( n \) approaches infinity. Convergence of a sequence means that as \( n \) becomes very large, the sequence approaches some finite limit.To analyze this sequence, observe the following:1. **Numerator Analysis**: - The term \( ne^{-n} \) in the numerator can be broken down as \( n \times e^{-n} \). - As \( n \to \infty \), \( e^{-n} \) (which is equivalent to \( \frac{1}{e^n} \)) decreases rapidly towards 0. - Meanwhile, \( n \) increases without bound. 2. **Denominator Analysis**: - The term \( n + e^{-n} \) in the denominator consists of two parts: \( n \) and \( e^{-n} \). - As \( n \to \infty \), the term \( e^{-n} \) becomes negligible in comparison to \( n \).3. **Behavior of Fraction**: - Since \( e^{-n} \to 0 \), both the numerator \( ne^{-n} \) and the additional term in the denominator \( e^{-n} \) in the given formula also approach 0. - Consequently, the numerator \( ne^{-n} \) decreases faster to 0 than the denominator due to its exponential decline.Considering these observations, the fraction \( \frac{ne^{-n}}{n + e^{-n}} \) simplifies for large \( n \) as follows:\[ \frac{ne^{-n}}{n + e^{-n}} \approx \frac{0}{n} = 0 \]Therefore, as \( n \) approaches infinity, the sequence \( \left\{ \frac{ne^{-n}}{n + e^{-n}} \right\}_{n=1}^{\infty} \) converges **Does the series converge? If so, what does it converge to? If not, why not?**\[ \sum_{n=0}^{\infty} \frac{2}{(-5)^n} \]To determine whether the series converges, check if the series forms a geometric series and evaluate its common ratio. If the common ratio is between -1 and 1 (exclusive), the series converges.### Analysis:We can rewrite the given series as:\[ \sum_{n=0}^{\infty} \frac{2}{(-5)^n} \] This series is in the form of a geometric series where \( a = 2 \) (the first term) and \( r = \frac{1}{-5} = -\frac{1}{5} \) (the common ratio).### Convergence Test for Geometric Series:A geometric series converges if \( |r| < 1 \). For our series:\[ |r| = \left| -\frac{1}{5} \right| = \frac{1}{5} < 1\]Since the common ratio \( \left| -\frac{1}{5} \right| \) is less than 1, the series converges.### Sum of the Convergent Series:The sum, \( S \), of an infinite geometric series can be found using the formula:\[ S = \frac{a}{1 - r} \]Substitute \( a = 2 \) and \( r = -\frac{1}{5} \):\[ S = \frac{2}{1 - \left( -\frac{1}{5} \right)} = \frac{2}{1 + \frac{1}{5}} = \frac{2}{\frac{6}{5}} = 2 \cdot \frac{5}{6} = \frac{10}{6} = \frac{5}{3} \]### Conclusion:The series converges, and its sum is \(\frac{5}{3}\).

1) The given sequence is; (ne-n/(n+e-n))infinity n=1. 2) The given series is 2/(-5)n.

Answered: Does the sequence converge? If so, what…

Does the sequence converge? If so, what is its limit? If not why not? ne-n nten n=1

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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2. Consider the sequence an given by 2n +1 n³ (a) Calculate the first four terms of the sequence (as integers or fractions). an = for n ≥ 1. (b) Is this sequence bounded above or below, or not bounded at all? Explain your answer in detail. Provide bounds if possible. (c) Does this sequence converge or diverge? If it converges, find the limit.

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