Radius and Interval of convergence

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Homework Help Overview

The discussion revolves around determining the radius and interval of convergence for the series Ʃ(((-1)^n)(x^n))/(n+1) from 0 to ∞. The original poster attempts to apply the ratio test and identifies the radius of convergence as 1, with an interval of (-1, 1). They are specifically examining the convergence at the endpoints of the interval.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster discusses the use of the alternating series test and expresses uncertainty about proving the inequality 1/(n+2) < 1/(n+1). They question whether there are alternative tests that could be applied, given the limitations of the ratio test.

Discussion Status

Participants are engaging with the original poster's concerns, suggesting starting points for proving the inequality and discussing properties of functions. There is an exploration of different approaches to demonstrate the necessary conditions for convergence, but no consensus has been reached on a definitive method.

Contextual Notes

The original poster mentions specific requirements from their professor regarding the proof of inequalities, indicating constraints on the discussion. There is also a reference to the challenges of applying the integral test to an alternating series.

Painguy
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Homework Statement


Ʃ(((-1)^n)(x^n))/(n+1) from 0 to ∞

Homework Equations


The Attempt at a Solution


I took applied the ratio test and got that lim|(x^(n+1))/(n+2) * (n+1)/(x^n)| =|x|
so that means for it to converge |x|<1 Radius of convergence is 1
My interval is (-1<x<1)
Now I check the endpoints

x=1
Ʃ(((-1)^n))/(n+1) from 0 to ∞
I want to use the alternating series test, but I have run into a little problem.
I know that the Lim 1/(n+1) =0 but how do I prove that 1/(n+2) < 1/(n+1) for all n.
I know its true, but my professor wants us to show that. I can show that -1<1 which is a true statement, but what does that say about the inequality in relation to n?

Is there a better test to do this? Ratio test just equals 1 so that tells me nothing. I can't really use the integral test for an alternating series since it would end up imaginary. If there is a better test could someone still show how to use the alternating series if possible?
 
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Usually you can assume certain properties of real numbers to be true. If your professor really expects you to prove such a trivial point, why don't you start with something "obviously" true, like n+2 > n+1 for all n?
 
Or, to start from the very obvious 2> 1. Add n to both sides, then divide both sides by ...
 
I set 1/(n+1) to a function f which I then siad was less than 0. when i took the derivative of f i obtained a negative function which is less than 0. I stated that since the derivative is less than zero then the function is decreasing and bn+1 < bn. Is that ok?
 

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