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

Use the Limit Comparison Test to determine the convergence or divergence of the series.
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}\left({n}^{2}+4\right)}$
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We have to find the given series is convergence or divergence.
Limit comparison test:
Suppose that we have two series $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$ and $\sum _{n=1}^{\mathrm{\infty }}{b}_{n}$ with ${a}_{n}\ge 0$ and ${b}_{n}>0$ for all n
Then, if $\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}=c$, with 0 then either both series convergent or both series divergent.
We have given
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}\left({n}^{2}+4\right)}$
Let $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}\left({n}^{2}+4\right)}$
Then, ${a}_{n}=\frac{1}{{n}^{2}\left({n}^{2}+4\right)}$
And suppose $\sum _{n=1}^{\mathrm{\infty }}{b}_{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{4}}$ another series.
Then, ${b}_{n}=\frac{1}{{n}^{4}}$
And $\sum _{n=1}^{\mathrm{\infty }}{b}_{n}$ is a p-series and p=4 so, it is convergent.
Now,
$\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{{a}^{2}\left({a}^{2}+4\right)}}{\frac{1}{{a}^{4}}}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{2}\left({n}^{2}+4\right)}×\frac{{n}^{4}}{1}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{{n}^{4}}{{n}^{4}\left(1+\frac{4}{{a}^{2}}\right)}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\left(1+\frac{4}{{a}^{2}}\right)}$
$=1$ (finite and non zero)
Since, series $\sum _{n=1}^{\mathrm{\infty }}{b}_{n}$ is convergent so, by comparison test series $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$ is convergent.
Therefore, given series is convergent.

Jeffrey Jordon