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)}

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asked 2020-10-23
Use the Limit Comparison Test to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{1}{n^2(n^2+4)}\)

Answers (1)

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.
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