# Question # Use the Limit Comparison Test to determine the convergence or divergence of the series. sum_{n=1}^inftyfrac{1}{n^2(n^2+4)}

Series
ANSWERED Use the Limit Comparison Test to determine the convergence or divergence of the series.
$$\sum_{n=1}^\infty\frac{1}{n^2(n^2+4)}$$ 2020-10-24
We have to find the given series is convergence or divergence.
Limit comparison test:
Suppose that we have two series $$\sum_{n=1}^\infty a_n$$ and $$\sum_{n=1}^\infty b_n$$ with $$a_n\geq0$$ and $$b_n>0$$ for all n
Then, if $$\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=c$$, with $$0 then either both series convergent or both series divergent. We have given \(\sum_{n=1}^\infty\frac{1}{n^2(n^2+4)}$$
Let $$\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{1}{n^2(n^2+4)}$$
Then, $$a_n=\frac{1}{n^2(n^2+4)}$$
And suppose $$\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty\frac{1}{n^4}$$ another series.
Then, $$b_n=\frac{1}{n^4}$$
And $$\sum_{n=1}^\infty b_n$$ is a p-series and p=4 so, it is convergent.
Now,
$$\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=\lim_{n\rightarrow\infty}\frac{\frac{1}{a^2(a^2+4)}}{\frac{1}{a^4}}$$
$$=\lim_{n\rightarrow\infty}\frac{1}{n^2(n^2+4)}\times\frac{n^4}{1}$$
$$=\lim_{n\rightarrow\infty}\frac{n^4}{n^4(1+\frac{4}{a^2})}$$
$$=\lim_{n\rightarrow\infty}\frac{1}{(1+\frac{4}{a^2})}$$
$$=1$$ (finite and non zero)
Since, series $$\sum_{n=1}^\infty b_n$$ is convergent so, by comparison test series $$\sum_{n=1}^\infty a_n$$ is convergent.
Therefore, given series is convergent.