Calc6.21

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Discuss the convergence or divergence of the series \sum _{{n=1}}^{{\infty }}{\frac  {2n^{2}}{4n^{4}+8n^{3}+3}}.

Since the degree of the denominator is 2 greater than the degree of the numerator, we can most likely compare this to the series \sum _{{n=1}}^{{\infty }}{\frac  {1}{n^{2}}}. First, let us manipulate our series a bit.

{\frac  {2n^{2}}{4n^{4}+8n^{3}+3}}={\frac  {1}{2n^{2}+4n+{\frac  {3}{2n^{2}}}}}\,

We can divide by n^{2} since n is never 0. Now, since {\frac  {1}{2n^{2}+4n+{\frac  {3}{2n^{2}}}}}<{\frac  {1}{n^{2}}}, and because the terms of both series are all positive, we can use the direct comparison test to conclude that our series, \sum _{{n=1}}^{{\infty }}{\frac  {2n^{2}}{4n^{4}+8n^{3}+3}}, converges.


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