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c) E(-1)" 00 Ln=1 3η

c) E(-1)" 00 Ln=1 3η

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question

Determine if each of the following series is absolutely convergent, conditionally convergent, or divergent

The image presents an infinite series for evaluation. The series is given by: \[ c) \sum_{n=1}^{\infty} (-1)^n \frac{n^3}{3^n} \] This represents an alternating series, where each term is of the form: \[ (-1)^n \frac{n^3}{3^n} \] The summation starts from \( n = 1 \) and continues indefinitely, denoted by the upper limit of infinity (\(\infty\)). The term \((-1)^n\) indicates that the series alternates between positive and negative terms. The expression \(\frac{n^3}{3^n}\) represents the term's magnitude as a function of \( n \), with \( n^3 \) being divided by \( 3^n \). This suggests that as \( n \) increases, the terms decrease in size due to the exponential growth of the denominator relative to the numerator.
Transcribed Image Text:The image presents an infinite series for evaluation. The series is given by: \[ c) \sum_{n=1}^{\infty} (-1)^n \frac{n^3}{3^n} \] This represents an alternating series, where each term is of the form: \[ (-1)^n \frac{n^3}{3^n} \] The summation starts from \( n = 1 \) and continues indefinitely, denoted by the upper limit of infinity (\(\infty\)). The term \((-1)^n\) indicates that the series alternates between positive and negative terms. The expression \(\frac{n^3}{3^n}\) represents the term's magnitude as a function of \( n \), with \( n^3 \) being divided by \( 3^n \). This suggests that as \( n \) increases, the terms decrease in size due to the exponential growth of the denominator relative to the numerator.
Expert Solution
Step 1

Let,    ∑n=1∞(-1)nn33n=∑n=1∞an

So,    an=(-1)nn33n

Replace n with n+1 in an,

an+1=(-1)n+1(n+1)33n+1=(-1)n.(-1)(n+1)33n.3

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