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Применяем радикальный признак Коши Так как предел имеет значение <1, то согласно признака ряд сходится.<br>
![\lim_{n \to \infty} \sqrt[n]{a_n}= \lim_{n \to \infty} \sqrt[n]{ (\frac{3n ^{2}+2 }{5+3n^{2} }) ^{n ^{3} } } = \\ =\lim_{n \to \infty} ((1-\frac{3}{5+3n ^{2} }) ^{ -\frac{5+3n^{2} }{-3} } ) ^{ \frac{-3n^{2} }{5+3n ^{2} } }=e ^{ \lim_{n \to \infty} \frac{-3n ^{2} }{5+3n ^{2} } }=e ^{-1}= \frac{1}{e}<1](https://tex.z-dn.net/?f=+%5Clim_%7Bn+%5Cto+%5Cinfty%7D++%5Csqrt%5Bn%5D%7Ba_n%7D%3D+%5Clim_%7Bn+%5Cto+%5Cinfty%7D++%5Csqrt%5Bn%5D%7B+%28%5Cfrac%7B3n+%5E%7B2%7D%2B2+%7D%7B5%2B3n%5E%7B2%7D+%7D%29+%5E%7Bn+%5E%7B3%7D+%7D+%7D+%3D++%5C%5C+%3D%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%28%281-%5Cfrac%7B3%7D%7B5%2B3n+%5E%7B2%7D+%7D%29+%5E%7B+-%5Cfrac%7B5%2B3n%5E%7B2%7D+%7D%7B-3%7D+%7D+%29+%5E%7B+%5Cfrac%7B-3n%5E%7B2%7D+%7D%7B5%2B3n+%5E%7B2%7D+%7D+%7D%3De+%5E%7B+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7B-3n+%5E%7B2%7D+%7D%7B5%2B3n+%5E%7B2%7D+%7D++%7D%3De+%5E%7B-1%7D%3D+%5Cfrac%7B1%7D%7Be%7D%3C1++++++ "\lim_{n \to \infty} \sqrt[n]{a_n}= \lim_{n \to \infty} \sqrt[n]{ (\frac{3n ^{2}+2 }{5+3n^{2} }) ^{n ^{3} } } = \\ =\lim_{n \to \infty} ((1-\frac{3}{5+3n ^{2} }) ^{ -\frac{5+3n^{2} }{-3} } ) ^{ \frac{-3n^{2} }{5+3n ^{2} } }=e ^{ \lim_{n \to \infty} \frac{-3n ^{2} }{5+3n ^{2} } }=e ^{-1}= \frac{1}{e}<1")