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(d) 3n 2n + 1 8 n=1

(d) 3n 2n + 1 8 n=1

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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Use definition of convergence to prove each of the following sequences converge. (Can you please do part D) thanks

The image presents two mathematical series: (c) \(\{2^{-n}\}_{n=1}^{\infty}\) This represents an infinite sequence where each term is the reciprocal of a power of 2. The terms are \(2^{-1}, 2^{-2}, 2^{-3}, \ldots\) (d) \(\left\{\frac{3n}{2n + 1}\right\}_{n=1}^{\infty}\) This denotes an infinite sequence where each term is the result of the fraction \(\frac{3n}{2n + 1}\). The sequence begins with substituting \(n = 1\), then \(n = 2\), and so on.
Transcribed Image Text:The image presents two mathematical series: (c) \(\{2^{-n}\}_{n=1}^{\infty}\) This represents an infinite sequence where each term is the reciprocal of a power of 2. The terms are \(2^{-1}, 2^{-2}, 2^{-3}, \ldots\) (d) \(\left\{\frac{3n}{2n + 1}\right\}_{n=1}^{\infty}\) This denotes an infinite sequence where each term is the result of the fraction \(\frac{3n}{2n + 1}\). The sequence begins with substituting \(n = 1\), then \(n = 2\), and so on.
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Step 1: Introduction of the given problem

We have to show that the given sequence open curly brackets fraction numerator 3 n over denominator 2 n plus 1 end fraction close curly brackets subscript n equals 1 end subscript superscript infinity is convergent.

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