How do you know if the series #(1+n+n^2)/(sqrt(1+(n^2)+n^6))# converges or diverges for (n=1 , ∞) ?

Question

Christopher Agresta · Accepted Answer

#((1+n+n^2)/sqrt(1+n^2+n^6))^2 = (1+2n+3n^2+2n^3+n^4)/(1+n^2+n^6)# #=1/n^2*(n^2+2n^3+3n^4+2n^5+n^6)/(1+n^2+n^6)# #=1/n^2*((1+n^2+n^6)+(2n^3+3n^4+2n^5-1))/(1+n^2+n^6)# #=1/n^2*(1+(2n^3+3n^4+2n^5-1)/(1+n^2+n^6))# #>1/n^2# for all #n >= 1# So #(1+n+n^2)/sqrt(1+n^2+n^6) > 1/n# And #sum_(n=1)^oo (1+n+n^2)/sqrt(1+n^2+n^6) > sum_(n=1)^oo 1/n# diverges.

Axel Alvarez · Answer

undefined To determine if the series $ \frac{1+n+n^2}{\sqrt{1+n^2+n^6}} $ converges or diverges as $ n $ approaches infinity, you can use the limit comparison test. First, find a series $ b_n $ that behaves similarly to the given series. Then, take the limit as $ n $ approaches infinity of the ratio of the two series. If the limit is a finite positive number, both series either converge or diverge. If the limit is zero or infinite, the behavior of the two series may differ.

In this case, we can compare the given series to the series $ b_n = \frac{n^2}{n^3} = \frac{1}{n} $, which is a p-series with $ p = 1 $. Taking the limit of the ratio $ \frac{a_n}{b_n} $ as $ n $ approaches infinity, where $ a_n $ is the given series, we get:

$$ \lim_{n \to \infty} \frac{\frac{1+n+n^2}{\sqrt{1+n^2+n^6}}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n(1+n+n^2)}{\sqrt{n^2(n^2+1+n^4)}} $$

Simplify the expression inside the limit:

$$ = \lim_{n \to \infty} \frac{n+n^2+n^3}{\sqrt{n^4+n^2+n^6}} $$

As $ n $ approaches infinity, $ n^3 $ grows faster than $ n $, $ n^2 $, and $ \sqrt{n^4+n^2+n^6} $. Therefore, we can ignore the terms involving $ n $ and $ n^2 $ in both the numerator and denominator:

$$ = \lim_{n \to \infty} \frac{n^3}{\sqrt{n^6}} = \lim_{n \to \infty} \frac{n^3}{n^3} = 1 $$

Since the limit is a finite positive number, and $ b_n = \frac{1}{n} $ is a convergent p-series, by the limit comparison test, the given series $ \frac{1+n+n^2}{\sqrt{1+n^2+n^6}} $ also converges.