Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the convergence or divergence of the series using other methods. sum_{n=1}^inftyfrac{n^2}{(n+1)(n^2+2)}

Rui Baldwin 2021-03-18 Answered
Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the convergence or divergence of the series using other methods.
\(\sum_{n=1}^\infty\frac{n^2}{(n+1)(n^2+2)}\)

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Expert Answer

StrycharzT
Answered 2021-03-19 Author has 12418 answers

Ratio test:
Let the series is \(\sum a_n\) then,
\(\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L\)
If \(L<1\), series is convergent.
If \(L>1\), series is divergent.
If \(L=1\), may be convergent or divergent.
Given that,
\(\sum_{n=1}^\infty\frac{n^2}{(n+1)(n^2+2)}\)
Here
\(a_n=\frac{n^2}{(n+1)(n^2+2)}\) so
\(a_{n+1}=\frac{(n+1)^2}{(n+2)((n+1)^2+2)}\)
Now apply the ratio test,
\(\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}(\frac{\frac{(n+1)^2}{(n+2)((n+1)^2+2)}}{\frac{n^2}{(n+1)(n^2+2)}})\)
\(\lim_{n\to\infty}\left(\frac{(n+1)^2}{(n+2)((n+1)^2+2)}\times\frac{(n+1)(n^2+2)}{n^2}\right)\)
\(\lim_{n\to\infty}\left(\frac{n^2(1+\frac{1}{n})^2}{n^3(1+\frac{2}{n})((1+\frac{1}{n})^2+\frac{2}{n^2})}\times\frac{n^3(1+\frac{1}{n})(1+\frac{2}{n^2})}{n^2}\right)\)
\(\lim_{n\to\infty}\left(\frac{(1+\frac{1}{n})^2}{(1+\frac{2}{n})((1+\frac{1}{n})^2+\frac{2}{n^2})}\times\frac{(1+\frac{1}{n})(1+\frac{2}{n^2})}{1}\right)\)
=1
Since \(\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L\), where \(L=1\). So ratio test fails.
Comparison test:
Let the series is \(\sum a_n\) then,
\(\lim_{n\to\infty}\frac{a_n}{b_n}=L\)
Where L is positive and finite then \(\sum a_n\) and \(\sum b_n\) are both convergent or divergent.
Given that,
\(\sum_{n=1}^\infty\frac{n^2}{(n+1)(n^2+2)}\)
Here \(a_n=\frac{n^2}{(n+1)(n^2+2)}\) so
\(b_n=\frac{n^2}{n^3}=\frac{1}{n}\) Now using the comparison test,
\(\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\left(\frac{\frac{n^2}{(n+1)(n^2+2)}}{\frac{1}{n}}\right)\)
\(=\lim_{n\to\infty}\left(\frac{n^2}{(n+1)(n^2+2)}\times n\right)\)
\(=\lim_{n\to\infty}\left(\frac{n^2}{n^3(1+\frac{1}{n})(1+\frac{2}{n^2})}\times n\right)\)
\(=\lim_{n\to\infty}\left(\frac{1}{(1+\frac{1}{n})(1+\frac{2}{n^2})}\right)\)
\(=1\)
Since \(=\lim_{n\to\infty}\frac{a_n}{b_n}=1\), which is positive and finite.
Since \(b_n=\frac{1}{n}\), so using p series test
[\(\sum_{n=1}^\infty\frac{1}{n^p}\) is convergent if \(p>1\) or divergent if \(p\leq1\)], \(b_n=\frac{1}{n}\) is divergent.
Since \(\lim_{n\to\infty}\frac{a_n}{b_n}=1\), which is positive and finite and \(\sum b_n\) is divergent, so using comparison test \(\sum a_n\) is also divergent
Hence \(\sum_{n=1}^\infty\frac{n^2}{(n+1)(n^2+2)}\) is divergent.

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