# How do you use the limit comparison test to determine if #Sigma (5n-3)/(n^2-2n+5)# from #[1,oo)# is convergent or divergent?

Concluding

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To use the limit comparison test, you need to find a series that behaves similarly to the given series but is easier to evaluate. In this case, let's choose the series ( \sum \frac{1}{n} ), a p-series with ( p = 1 ), which is known to diverge.

Then, we'll form the limit of the ratio between the two series:

[ \lim_{n \to \infty} \frac{\frac{5n-3}{n^2-2n+5}}{\frac{1}{n}} ]

Next, simplify the expression by multiplying both the numerator and the denominator by ( \frac{1}{n} ):

[ \lim_{n \to \infty} \frac{5-\frac{3}{n}}{1-\frac{2}{n}+\frac{5}{n^2}} ]

Now, evaluate the limit:

[ \lim_{n \to \infty} \frac{5-0}{1-0+0} = 5 ]

Since the limit is a finite non-zero number (in this case, 5), and the series ( \sum \frac{1}{n} ) diverges, the given series ( \sum \frac{5n-3}{n^2-2n+5} ) also diverges by the limit comparison test.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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