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. ∑k=3∞2k2k2−k−2

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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.  ∑k=3∞2k2k2−k−2

curwyrm · Accepted Answer

The given series is  ∑k=3∞2k2k2−k−2  The series ∑k=3∞2k2k2−k−2 can be simplified as follows.  ∑k=3∞2k2k2−k−2=∑k=3∞2k2−k−2k2  ∑k=3∞21−1k−2k2  ∑k=3∞ak where ak=f(k)  Consider the function f(k)=21−1k−2k2  Here 21−1k−2k2≠0 for all values of k in the interval [3,∞), so this function is continuous of k  Put 3 for k in f(k)=21−1k−2k2  f(3)=21−13−232  f(3)=21−13−29  =92  Put 4 for k in f(k)=21−1k−2k2  f(4)=21−14−216  =165  Put 5 for k in f(k)=21−1k−2k2  f(5)=21−15−252  f(5)=21−15−225  =259  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.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

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}

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