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.

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.