Determine if the following series converges or diverges. If it is a converging geometric or telescoping series, or ca be written as one, provide what the series converges to. sum_{k=2}^inftyfrac{k}{k^3-17}

Bergen 2021-03-11 Answered
Determine if the following series converges or diverges. If it is a converging geometric or telescoping series, or ca be written as one, provide what the series converges to.
\(\displaystyle{\sum_{{{k}={2}}}^{\infty}}{\frac{{{k}}}{{{k}^{{3}}-{17}}}}\)

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Expert Answer

Maciej Morrow
Answered 2021-03-12 Author has 4404 answers
The series is \(\displaystyle{\sum_{{{k}={2}}}^{\infty}}{\frac{{{k}}}{{{k}^{{3}}-{17}}}}\)
Limit Comparison test: Suppose \(\displaystyle\sum{a}_{{n}}\) and \(\displaystyle\sum{b}_{{n}}\) are two series and \(\displaystyle\lim_{{{n}\to\infty}}{\frac{{{a}_{{n}}}}{{{b}_{{n}}}}}={c}\). If c is a finite and positive number then both the series converge or diverge together.
Let's take the second series \(\displaystyle\sum{b}_{{n}}=\sum{\frac{{{1}}}{{{n}^{{2}}}}}\) which is known as convergent by p-series.
\(\displaystyle\sum{a}_{{n}}=\sum{\frac{{{n}}}{{{n}^{{3}}-{17}}}}\quad\sum{b}_{{n}}=\sum{\frac{{{1}}}{{{n}^{{2}}}}}\)
\(\displaystyle\lim_{{{n}\to\infty}}{\frac{{{a}_{{n}}}}{{{b}_{{n}}}}}=\lim_{{{n}\to\infty}}{\frac{{{\frac{{{n}}}{{{n}^{{3}}-{17}}}}}}{{{\frac{{{1}}}{{{n}^{{2}}}}}}}}=\lim_{{{n}\to\infty}}{\frac{{{n}}}{{{n}^{{3}}{\left({1}-{\frac{{{17}}}{{{n}^{{3}}}}}\right)}}}}\times{\frac{{{n}^{{2}}}}{{{1}}}}=\lim_{{{n}\to\infty}}{\frac{{{1}}}{{{\left({1}-{\frac{{{17}}}{{{n}^{{3}}}}}\right)}}}}={1}\)
As the ration of the two series is 1 which is finite and positive, using Limit Comparison test the series \(\displaystyle\sum{a}_{{n}}=\sum{\frac{{{n}}}{{{n}^{{3}}-{17}}}}\) converges as the series \(\displaystyle\sum{b}_{{n}}=\sum{\frac{{{1}}}{{{n}^{{2}}}}}\) convergent.
Hence \(\displaystyle{\sum_{{{k}={2}}}^{\infty}}{\frac{{{k}}}{{{k}^{{3}}-{17}}}}\) is convergent.
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