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Comparison test to determine convergence

  • Thread starter masonm127
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  • #1
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Use a comparison test to determine whether the series [itex]\sum[/itex] (n+1)/(n[itex]^{2}[/itex]+n+1) diverges or converges.

I started out by simplifying the series to 1/n+1 and then from there I compared it to 1/n, which converges. 1/n is greater than 1/n+1 so based on the comparison test, the original series should also converge, is this correct? I also tried a limit comparison test and got n/n+1 which equals 2 which would mean that both Ʃa and Ʃb converge. I am kind of shady on my series and and getting very confused with this question.
thanks!

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  • #2
You could use this inequality [itex]n^2+n+1\le n^2+2n+1= (n+1)^2\quad\forall n\in\mathbb{N}[/itex] so:

[itex]\frac{1}{(n+1)^2}\le \frac{1}{n^2+n+1}[/itex]

multiplying both sides for [itex]n+1[/itex], we have:

[itex]\frac{1}{n+1}\le \frac{n+1}{n^2+n+1}[/itex]

So

[itex]\sum_{n=0}^\infty\frac{1}{n+1}\le \sum_{n=0}^\infty\frac{n+1}{n^2+n+1}[/itex]

but
[tex]\sum_{n=0}^{\infty}\frac{1}{n+1}=\sum_{m=1}^{\infty}\frac{1}{m}=\infty[/tex]

therefore [itex]\sum_{n=0}^\infty\frac{n+1}{n^2+n+1}=\infty[/itex]
 
Last edited:
  • #3
HallsofIvy
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First, what do you mean by "simplified the series to 1/n+ 1". "Simplifying" normally reducing to something equal by canceling, say. That is not the case here. Second, the series [itex]\sum (1/n)[/itex] does NOT converge.
 

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