# 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}

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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Maciej Morrow
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.