Use the Limit Comparison Test to determine the convergence or divergence of the series. sum_{n=1}^inftyfrac{2n^2-1}{3n^5+2n+1}

Kyran Hudson 2021-03-08 Answered
Use the Limit Comparison Test to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{2n^2-1}{3n^5+2n+1}\)

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Expert Answer

irwchh
Answered 2021-03-09 Author has 10451 answers

We have to check the given series
\(\sum_{n=1}^\infty\frac{2n^2-1}{3n^5+2n+1}\) is convergent or divergent using the limit comparision test.
According to limit comparision test if two series \(\sum_{n=1}^\infty a_n\) and \(\sum_{n=1}^\infty b_n\) with \(a_n>0,b_n>0\) for all n.
Then if \(\lim_{n\to\infty}\frac{a_n}{b_n}=c\) with 0 then either both series converges or both series divergent.
Let \(\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty a_n\frac{2n^2-1}{3n^5+2n+1}\)
Take common highest power n from denominator and numerator we get
\(\sum_{n=1}^\infty\frac{2n^2-1}{3n^5+2n+1}=\sum_{n=1}^\infty\frac{n^2(2-\frac{1}{n^2})}{n^5(3+\frac{2}{n^4}+\frac{1}{n^5})}\)
\(=\sum_{n=1}^\infty\frac{(2-\frac{1}{n^2})}{n^3(3+\frac{2}{n^4}+\frac{1}{n^5})}\)
So, \(\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{(2-\frac{1}{n^2})}{n^3(3+\frac{2}{n^4}+\frac{1}{n^5})}\)
Let another series \(b_n=\frac{1}{n^3}\)
\(b_n\) is convergent p-series since p=3
Now,
\(\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac{(2-\frac{1}{n^2})}{n^3(3+\frac{2}{n^4}+\frac{1}{n^5})}}{\frac{1}{n^3}}\)
\(=\lim_{n\to\infty}\frac{(2-\frac{1}{n^2})}{(3+\frac{2}{n^4}+\frac{1}{n^5})}\)
\(=\frac{2-0}{3+0+0}\)
\(=\frac23\)
\(=\lim_{n\to\infty}\frac{a_n}{b_n}=\frac{2}{3}\)
which is finite and positive.
Therefore we can conclude by limit comparison test series \(\sum_{n=1}^\infty a_n\) will be convergent.
So given series is convergent.

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