Identify a convergence test for the following series. If necessary, explain how to simplify or rewrite the series before applying the convergence test. You do not need to carry out the convergence test. sum_{k=3}^inftyfrac{2k^2}{k^2-k-2}

Question
Series
asked 2020-10-23
Identify a convergence test for the following series. If necessary, explain how to simplify or rewrite the series before applying the convergence test. You do not need to carry out the convergence test.
\(\sum_{k=3}^\infty\frac{2k^2}{k^2-k-2}\)

Answers (1)

2020-10-24
The given series is \(\sum_{k=3}^\infty\frac{2k^2}{k^2-k-2}\)
The series \(\sum_{k=3}^\infty\frac{2k^2}{k^2-k-2}\) can be simplified as follows.
\(\sum_{k=3}^\infty\frac{2k^2}{k^2-k-2}=\sum_{k=3}^\infty\frac{2}{\frac{k^2-k-2}{k^2}}\)
\(\sum_{k=3}^\infty\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)
\(\sum_{k=3}^\infty a_k\) where \(a_k=f(k)\)
Consider the function \(f(k)=\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)
Here \(\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\ne0\) for all values of k in the interval \([3,\infty)\), so this function is continuous of k
Put 3 for k in \(f(k)=\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)
\(f(3)=\frac{2}{1-\frac{1}{3}-\frac{2}{3^2}}\)
\(f(3)=\frac{2}{1-\frac{1}{3}-\frac{2}{9}}\)
\(=\frac92\)
Put 4 for k in \(f(k)=\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)
\(f(4)=\frac{2}{1-\frac{1}{4}-\frac{2}{16}}\)
\(=\frac{16}{5}\)
Put 5 for k in \(f(k)=\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)
\(f(5)=\frac{2}{1-\frac{1}{5}-\frac{2}{5^2}}\)
\(f(5)=\frac{2}{1-\frac{1}{5}-\frac{2}{25}}\)
\(=\frac{25}{9}\)
It is observed that the function is a deceasing function,
Therefore, it can be concluded that the Integral test is used to determine the convergence of the given series.
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