Comparison test to determine convergence

In summary, the conversation is discussing how to determine whether the series \sum (n+1)/(n^{2}+n+1) converges or diverges. The speaker suggests using a comparison test and a limit comparison test to analyze the series. They also provide an inequality that shows the original series is less than or equal to another series, which can help determine convergence. The conversation ends with confusion and uncertainty about the series.
  • #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]
 
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  • #3
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.
 

What is the comparison test method used for?

The comparison test is used to determine the convergence or divergence of an infinite series by comparing it to a known series whose convergence or divergence is already known.

How is the comparison test performed?

The comparison test involves comparing the infinite series in question to a known series with similar properties. If the known series converges, then the infinite series also converges. If the known series diverges, then the infinite series also diverges.

What is the role of the comparison series in the test?

The comparison series serves as a benchmark for determining the convergence or divergence of the infinite series. It allows us to make a comparison and draw a conclusion about the infinite series in question.

What are the different types of comparison tests?

There are three types of comparison tests: the direct comparison test, the limit comparison test, and the integral comparison test. Each test has its own specific conditions and criteria for determining convergence or divergence.

What are the limitations of the comparison test?

The comparison test can only be used for infinite series with non-negative terms. It also requires the comparison series to be known and have similar properties as the series in question. Additionally, the comparison test does not provide information about the actual value of the series, only its convergence or divergence.

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