Limit of Series for Infinity Sum: Basic Comparison Test

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of the infinite series given by the expression (2n+1) / (5n+1). Participants are exploring the application of the basic comparison test and other methods to analyze the series.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss splitting the fraction to analyze its behavior as n approaches infinity, questioning whether the limit of the sequence approaches zero, which is a condition for convergence. There are also considerations about dividing both the numerator and denominator by n to simplify the expression.

Discussion Status

The discussion includes various approaches to understanding the convergence of the series, with some participants suggesting that the limit of the sequence not approaching zero indicates divergence. There is no explicit consensus on the necessity of using specific tests, but some guidance is provided regarding the n-th term test for divergence.

Contextual Notes

Participants note that the original question specifically asks for the determination of convergence or divergence, and there is mention of the answer being divergence. The discussion reflects a focus on the conditions required for convergence without resolving the inquiry definitively.

teng125
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for infinity sum n=1 (2n+1) / (5n+1)

how to find the converges or diverges??
is it suitable to use basic comparison test??pls show me
 
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If we split the fraction, we get

[tex]\frac{{2n + 1}}{{5n + 1}} = \frac{2}{5} + \frac{3}{{25n + 5}}[/tex]

As you can see, the second term will go to 0 if n tends to infinity but there's a remaining term 2/5. In other words: the limit of the associated sequence isn't 0 (for n going to infinity) and this is a necessary (though not sufficient) condition for the convergence of the series.
 
Or: divide both numerator and denominator by n:
[tex]\frac{2n+1}{5n+1}= \frac{2+\frac{1}{n}}{5+\frac{1}{n}}[/tex]

Now what does 1/n go to as n goes to infinity?
 
but the question ask for diverges or converges??and the answer is diverges
 
When a series doesn't converge, it diverges.
 
is it necessary to use any rules such as the one i mentioned??or just to perform the steps showed above??
 
I don't think those are necessary, if the limit of the (positive) sequence doesn't go to 0 (which is easy to check without using those other tests), then the (positive) series doesn't converge.
 
teng the n-th term test for divergence is an easy check for this one.
 

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