Question

Use the Limit Comparison Test to prove convergence or divergence of the infinite series. sum_{n=1}^inftyfrac{e^n+n}{e^{2n}-n^2}

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asked 2021-01-10
Use the Limit Comparison Test to prove convergence or divergence of the infinite series.
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{e}^{{n}}+{n}}}{{{e}^{{{2}{n}}}-{n}^{{2}}}}}\)

Answers (1)

2021-01-11
We will use the Comparison Test for convergence of infinite series.
Given infinite series is:
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{e}^{{n}}+{n}}}{{{e}^{{{2}{n}}}-{n}^{{2}}}}}\)
Let us consider
\(\displaystyle{a}_{{n}}={\frac{{{e}^{{n}}+{n}}}{{{e}^{{{2}{n}}}-{n}^{{2}}}}}\)
\(\displaystyle{b}_{{n}}={\frac{{{1}}}{{{e}^{{n}}}}}\)
Then we get,
\(\displaystyle{\frac{{{a}_{{n}}}}{{{b}_{{n}}}}}={\frac{{{\frac{{{e}^{{n}}+{n}}}{{{e}^{{{2}{n}}}-{n}^{{2}}}}}}}{{{\frac{{{1}}}{{{e}^{{n}}}}}}}}={e}^{{n}}{\left({\frac{{{e}^{{n}}+{n}}}{{{e}^{{{2}{n}}}-{n}^{{2}}}}}\right)}={\frac{{{e}^{{{2}{n}}}+\ne^{{n}}}}{{{e}^{{{2}{n}}}-{n}^{{2}}}}}={\frac{{{1}+{\frac{{{n}}}{{{e}^{{n}}}}}}}{{{1}-{\frac{{{n}^{{2}}}}{{{e}^{{{2}{n}}}}}}}}}\)
\(\displaystyle\lim_{{{n}\to\infty}}{\left({\frac{{{a}_{{n}}}}{{{b}_{{n}}}}}\right)}=\lim_{{{n}\to\infty}}{\left({\frac{{{1}+{\frac{{{n}}}{{{e}^{{n}}}}}}}{{{1}-{\frac{{{n}^{{2}}}}{{{e}^{{{2}{n}}}}}}}}}\right)}={1}\)
Since we can prove by using Integral Test that the infinite series \(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{b}_{{n}}={\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{e}^{{n}}}}}\) is convergent,
Therefore the given infinite series \(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{a}_{{n}}={\sum_{{{n}={1}}}^{\infty}}{\frac{{{e}^{{n}}+{n}}}{{{e}^{{{2}{n}}}-{n}^{{2}}}}}\) is also convergent by Comparison Test.
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