Chuỗi số nào sau đây hội tụ?

$\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n^4+1}$

$\sum _{n=1}^{\infty }\frac{{(-1)}^n}{3^n}$

\({\rm{d}}y = 2\left( {x - 1} \right){\rm{d}}x\)

\({\rm{d}}y = {\left( {x - 1} \right)^2}{\rm{d}}x\)

\({\rm{d}}y = 2\left( {x - 1} \right)\)

\({\rm{d}}y = 2\left( {x - 1} \right){\rm{d}}x\)

\(dy = (3{x^2} - 4x)dx\)

\(dy = (3{x^2} + x)dx\)

\(dy = (3{x^2} + 2x)dx\)

\(dy = (3{x^2} + 4x)dx\)

\(dy = \frac{3}{{\sqrt {3x + 2} }}dx\)

\(dy = \frac{1}{{2\sqrt {3x + 2} }}dx\)

\(dy = \frac{1}{{\sqrt {3x + 2} }}dx\)

\(dy = \frac{3}{{2\sqrt {3x + 2} }}dx\)

\(dy = (1 + {\tan ^2}2x)dx\)

\(dy = (1 - {\tan ^2}2x)dx\)

\(dy = 2(1 - {\tan ^2}2x)dx\)

\(dy = 2(1 + {\tan ^2}2x)dx\)

\({\rm{d}}f(x) = \frac{{ - \sin 4x}}{{2\sqrt {1 + {{\cos }^2}2x} }}{\rm{d}}x\)

\({\rm{d}}f(x) = \frac{{ - \sin 4x}}{{\sqrt {1 + {{\cos }^2}2x} }}{\rm{d}}x\)

\({\rm{d}}f(x) = \frac{{\cos 2x}}{{\sqrt {1 + {{\cos }^2}2x} }}{\rm{d}}x\)

\({\rm{d}}f(x) = \frac{{ - \sin 2x}}{{2\sqrt {1 + {{\cos }^2}2x} }}{\rm{d}}x\)

\({\rm{d}}y = \frac{{{\rm{d}}x}}{{{{\left( {x - 1} \right)}^2}}}\)

\({\rm{d}}y = \frac{{3{\rm{d}}x}}{{{{\left( {x - 1} \right)}^2}}}\)

\({\rm{d}}y = \frac{{ - 3{\rm{d}}x}}{{{{\left( {x - 1} \right)}^2}}}\)

\({\rm{d}}y = - \frac{{{\rm{d}}x}}{{{{\left( {x - 1} \right)}^2}}}\)

\({\rm{d}}y = - \frac{{{x^2} - 2x - 2}}{{{{(x - 1)}^2}}}{\rm{d}}x\)

\({\rm{d}}y = \frac{{2x + 1}}{{{{(x - 1)}^2}}}{\rm{d}}x\)

\({\rm{d}}y = - \frac{{2x + 1}}{{{{(x - 1)}^2}}}{\rm{d}}x\)

\({\rm{d}}y = \frac{{{x^2} - 2x - 2}}{{{{(x - 1)}^2}}}{\rm{d}}x\)

\({\rm{d}}y = \left( { - \cos x + 3\sin x} \right){\rm{d}}x\)

\({\rm{d}}y = \left( { - \cos x - 3\sin x} \right){\rm{d}}x\)

\({\rm{d}}y = \left( {\cos x + 3\sin x} \right){\rm{d}}x\)

\({\rm{d}}y = - \left( {\cos x + 3\sin x} \right){\rm{d}}x\)

\[{\rm{d}}y = -\sin 2x\,{\rm{d}}x\]

\[{\rm{d}}y = \sin 2x\,{\rm{d}}x\]

\[{\rm{d}}y = \sin x\,{\rm{d}}x\]

\[{\rm{d}}y = {\rm{2cos}}x\,{\rm{d}}x\]