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?

Question

Ezekiel Alexander · Accepted Answer

#sum_(n=1)^oo (5n-3)/(n^2-2n+5)# is divergent

#(5n-3)/(n^2-2n+5+7n+5/4) < (5n-3)/(n^2-2n+5)# but #(5n-3)/(n^2-2n+5+7n+5/4) = (5n-3)/(n+5/2)^2=5/(n+5/2)-(31/2)/(n+5/2)^2# We know that #sum_(n=1)^oo1/n^2# is convergent and #1/(n+5/2)^2 < 1/n^2# so #sum_(n=1)^oo(31/2)/(n+5/2)^2# is convergent. Now examining #sum_(n=1)^oo5/(n+5/2) = sum_(n=3)^oo(5/(n+1/2)) gt 5sum_(n=4)^oo1/n# and we know that #sum_(k=4)^oo 1/n# is divergent. Concluding #sum_(n=1)^oo (5n-3)/(n^2-2n+5)# is divergent

William Abdalla · Answer

undefined 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.