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Question Number 102545 by 175mohamed last updated on 09/Jul/20

Answered by mathmax by abdo last updated on 13/Jul/20

$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{n}}^{1,2}-1}{{\mathrm{n}}^{2,3}+1}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{n}}^{\frac{6}{5}}-1}{{\mathrm{n}}^{\frac{23}{10}}+1}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{U}}_{\mathrm{n}}$$$${\mathrm{U}}_{\mathrm{n}}\sim \frac{{\mathrm{n}}^{\frac{6}{5}}}{{\mathrm{n}}^{\frac{23}{10}}}=\frac{1}{{\mathrm{n}}^{\frac{23}{10}-\frac{6}{5}}}=\frac{1}{{\mathrm{n}}^{\frac{23-12}{10}}}=\frac{1}{{\mathrm{n}}^{\frac{11}{10}}}\mathrm{and}\mathrm{\Sigma }\frac{1}{{\mathrm{n}}^{\frac{11}{10}}}\mathrm{is}\mathrm{convergent}(\frac{11}{10}>1)\Rightarrow \mathrm{\Sigma }{\mathrm{U}}_{\mathrm{n}}\mathrm{converges}$$

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