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

Question
Series
Use the Limit Comparison Test to determine the convergence or divergence of the series.
$$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{n}^{{{k}-{1}}}}}{{{n}^{{k}}+{1}}}},{k}{>}{2}$$

2020-11-09
Given: $$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{n}^{{{k}-{1}}}}}{{{n}^{{k}}+{1}}}},{k}{>}{2}$$
Explanation:
Let $$\displaystyle{a}_{{n}}={\frac{{{n}^{{{k}-{1}}}}}{{{n}^{{k}}+{1}}}}$$ then $$\displaystyle{a}_{{n}}{>}{0}\forall{n}\geq{1}$$
Let $$\displaystyle{b}_{{n}}={\frac{{{1}}}{{{n}}}}$$ then $$\displaystyle{b}_{{n}}{>}{0}$$
$$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{b}_{{n}}={\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{n}}}}$$ is a harmonic series, so it is divergent
Now, $$\displaystyle\lim_{{{n}\to\infty}}{\frac{{{a}_{{n}}}}{{{b}_{{n}}}}}=\lim_{{{n}\to\infty}}{\frac{{{\frac{{{n}^{{{k}-{1}}}}}{{{n}^{{k}}+{1}}}}}}{{{\frac{{{1}}}{{{n}}}}}}}$$
$$\displaystyle=\lim_{{{n}\to\infty}}{\frac{{{n}^{{k}}}}{{{n}^{{k}}+{1}}}}$$
$$\displaystyle=\lim_{{{n}\to\infty}}{\frac{{{1}}}{{{1}+{\frac{{{1}}}{{{n}^{{k}}}}}}}}$$
$$\displaystyle={\frac{{{1}}}{{{1}+{0}}}}$$
$$\displaystyle={1}$$
Since $$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{n}}}}$$ is divergent

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