Find the limits of the sequences \sum_n^\infty=1\frac{n\times\ln(n)}{n^{2}+1} \sum_n^\infty=1\frac{\sin(n)}{n}

bmgf3m 2022-01-07 Answered
Find the limits of the sequences
\(\displaystyle{\sum_{{n}}^{\infty}}={1}{\frac{{{n}\times{\ln{{\left({n}\right)}}}}}{{{n}^{{{2}}}+{1}}}}\)
\(\displaystyle{\sum_{{n}}^{\infty}}={1}{\frac{{{\sin{{\left({n}\right)}}}}}{{{n}}}}\)

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Expert Answer

Laura Worden
Answered 2022-01-08 Author has 2208 answers
Consider the series
\(\displaystyle{\sum_{{n}}^{\infty}}={1}{\frac{{{n}\times{\ln{{\left({n}\right)}}}}}{{{n}^{{{2}}}+{1}}}}\)
\(\displaystyle{\sum_{{n}}^{\infty}}={1}{\frac{{{\sin{{\left({n}\right)}}}}}{{{n}}}}\)
By applying the Squeeze Theorem, we have
\(\displaystyle\lim_{{{n}\rightarrow\infty}}{\frac{{{\sin{{\left({n}\right)}}}}}{{{n}}}}={0}\)
\(\displaystyle\lim_{{{n}\rightarrow\infty}}{\frac{{{n}\times{\ln{{\left({n}\right)}}}}}{{{n}^{{{2}}}+{1}}}}\)
\(\displaystyle=\lim_{{{n}\rightarrow\infty}}{\frac{{{\frac{{{\ln{{\left({n}\right)}}}}}{{{n}}}}}}{{{1}+{\frac{{{1}}}{{{n}^{{{2}}}}}}}}}\)
This implies
\(\displaystyle\lim_{{{n}\rightarrow\infty}}{\frac{{{n}\times{\ln{{\left({n}\right)}}}}}{{{n}^{{{2}}}+{1}}}}\)
\(\displaystyle={\frac{{\lim_{{{n}\rightarrow\infty}}{\frac{{{\ln{{\left({n}\right)}}}}}{{{n}}}}}}{{\lim_{{{n}\rightarrow\infty}}{1}+{\frac{{{1}}}{{{n}^{{{2}}}}}}}}}\)
\(\displaystyle={\frac{{{0}}}{{{1}}}}={0}\)
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movingsupplyw1
Answered 2022-01-09 Author has 5411 answers
O, god, thanks
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