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
asked 2020-11-08
Use the Limit Comparison Test to determine the convergence or divergence of the series.
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{n}^{{{k}-{1}}}}}{{{n}^{{k}}+{1}}}},{k}{>}{2}\)

Answers (1)

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
0

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