How do you use the limit comparison test for #sum (2x^4)/(x^5+10)# n=1 to #n=oo#?

Question

Emilia Aguilar · Accepted Answer

#color(red)(sum_(n=1)^∞ (2x^4)/(x^5-10)" is divergent")#. #sum_(n=1)^∞ (2x^4)/(x^5+10)# The limit comparison test states that if #a_n# and #b_n# are series with positive terms and if #lim_(n→∞) (a_n)/(b_n)# is positive and finite, then either both series converge or both diverge. Let #a_n = (2x^4)/(x^5+10)# Let's think about the end behaviour of #a_n#. For large #n#, the denominator #x^5+10# acts like #x^5#. So, for large #n#, #a_n# acts like #(2x^4)/x^5 = 2/x#. Let #b_n= 1/x# Then #lim_(n→∞)(a_n/b_n) = lim_(n→∞)( ((2x^4)/(x^5+10))/(1/x)) = lim_(n→∞)( (2x^4×x)/(x^5+10)) = lim_(n→∞)( (2x^5)/(x^5+10)) = lim_(n→∞)( 2/(1-1/x^5)) =2# The limit is both positive and finite, so either #a_n# and #b_n# are both divergent or both are convergent. But #b_n= 1/x# is divergent, so #a_n = x^4/(x^5-10)# is also divergent.

Emma Aldridge · Answer

undefined To use the limit comparison test for the series $ \sum \frac{2x^4}{x^5 + 10} $ as $ n $ approaches infinity, we need to compare it with a known series. Let's denote the given series as $ a_n = \frac{2x^4}{x^5 + 10} $.

First, we find a series $ b_n $ that is easier to evaluate but behaves similarly to $ a_n $. In this case, we can choose $ b_n = \frac{1}{x} $ as it is simpler and has similar terms in the denominator.

Next, we take the limit as $ n $ approaches infinity of the ratio $ \frac{a_n}{b_n} $:

$$
\lim_{{n \to \infty}} \frac{a_n}{b_n} = \lim_{{n \to \infty}} \frac{\frac{2x^4}{x^5 + 10}}{\frac{1}{x}} = \lim_{{n \to \infty}} \frac{2x^5}{x^5 + 10}
$$

Solving the limit, we get:

$$
\lim_{{n \to \infty}} \frac{2x^5}{x^5 + 10} = \frac{2}{1} = 2
$$

Since the limit is a finite positive number, we conclude that the series $ \sum \frac{2x^4}{x^5 + 10} $ behaves similarly to the series $ \sum \frac{1}{x} $ as $ n $ approaches infinity.

Now, we can apply the limit comparison test. Since $ \sum \frac{1}{x} $ is a known divergent series (harmonic series), and $ \sum \frac{2x^4}{x^5 + 10} $ behaves similarly to it, we conclude that $ \sum \frac{2x^4}{x^5 + 10} $ diverges as well.