# 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}

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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Laura Worden
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
$$\displaystyle\lim_{{{n}\rightarrow\infty}}{\frac{{{n}\times{\ln{{\left({n}\right)}}}}}{{{n}^{{{2}}}+{1}}}}$$
$$\displaystyle={\frac{{\lim_{{{n}\rightarrow\infty}}{\frac{{{\ln{{\left({n}\right)}}}}}{{{n}}}}}}{{\lim_{{{n}\rightarrow\infty}}{1}+{\frac{{{1}}}{{{n}^{{{2}}}}}}}}}$$
$$\displaystyle={\frac{{{0}}}{{{1}}}}={0}$$
movingsupplyw1
O, god, thanks