Convergence of 2 + and 1 - alternating series

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Homework Help Overview

The problem involves determining the convergence or divergence of the series: 1 + 1/2 - 1/3 + 1/4 + 1/5 - 1/6 + ... Participants are exploring the properties of this alternating series and its transformation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the Leibniz rule and question its applicability to the series. There is an exploration of rewriting the series in terms of a function f(n) and examining its behavior as n approaches infinity. Some participants suggest using a comparison test to analyze the series further.

Discussion Status

The discussion is active, with participants questioning the limits of f(n) and its implications for convergence. There is a suggestion to use a comparison test, and some participants express uncertainty about the clarity of their comparisons. The conversation reflects a collaborative effort to refine reasoning and approaches.

Contextual Notes

Participants note that the series does not sum over all n but only specific values (n=2, 5, 8, etc.), which may affect the convergence analysis. There is also mention of the limitations of the ratio test in this context.

benf.stokes
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Homework Statement



The problem asks you if the series:
1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}...

converges or diverges

Homework Equations


The Attempt at a Solution



I tried to apply the Leibniz rule but I realized it can't be applied. Is there a transformation of this series that enables the use of Leibniz rule?
 
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If you write f(n)=1/n-1/(n+1)+1/(n+2), then the series becomes 1+f(2)+f(5)+f(8)+... Can you say something about the behavior of f(n) that might make you think it converges or diverges?
 
lim n tend to infinity of f(n) is different from zero and thus the series does not converge?
 
benf.stokes said:
lim n tend to infinity of f(n) is different from zero and thus the series does not converge?

Why do you think lim n->infinity f(n) isn't zero??
 
Sorry. I misread your post. It is indeed zero. How then is this rewriting useful? f(n) is always positive and the ratio test is useless as it gives r=1
 
benf.stokes said:
Sorry. I misread your post. It is indeed zero. How then is this rewriting useful? f(n) is always positive and the ratio test is useless as it gives r=1

Try a comparison test. Can you make a simple estimate of the size of f(n) for large n? Combine the terms in f(n).
 
f(n)=\frac{n^2+2*n+6}{n*(n+1)*(n+2)}=\frac{n+1}{n*(n+2)}+\frac{5}{n*(n+1)*(n+2)} > \frac{n+1}{n*(n+2)}>\frac{1}{n}

and thus by the comparison test the series diverges?
 
benf.stokes said:
f(n)=\frac{n^2+2*n+6}{n*(n+1)*(n+2)}=\frac{n+1}{n*(n+2)}+\frac{5}{n*(n+1)*(n+2)} > \frac{n+1}{n*(n+2)}>\frac{1}{n}

and thus by the comparison test the series diverges?

I think your comparisons might be a bit sloppy. But you've got an n^2 in the numerator and an n^3 in the denominator, so you can certainly show f(n)>C/n for some value of C. You also have a be a little careful because the series doesn't sum over all n. Only over n=2,5,8,etc. So you'll need an argument of show that series diverges as well. But sure, once you have f(n)>C/n, the series diverges.
 
Thanks for everything . One last thing though: how should I do the comparison so that it's less sloppy?
 
Last edited:

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