比值审敛法










无穷级数

ζ(s)=∑k=1∞1ks{displaystyle zeta (s)=sum _{k=1}^{infty }{frac {1}{k^{s}}}}zeta (s)=sum _{k=1}^{infty }{frac {1}{k^{s}}}


无穷级数











比值审敛法是判别级数敛散性的一种方法,又称为达朗贝尔判别法D'Alembert's test)。




目录






  • 1 定理


  • 2 例子


    • 2.1 收敛


    • 2.2 发散


    • 2.3 不能确定




  • 3 参见





定理


n=1∞un{displaystyle sum _{n=1}^{infty }u_{n}}sum _{n=1}^{infty }u_{n}为正项级数,其中每一項皆為非 0 的實數或複數,如果


limn→|un+1un|=ρ{displaystyle lim _{nto infty }left|{frac {u_{n+1}}{u_{n}}}right|=rho }{displaystyle lim _{nto infty }left|{frac {u_{n+1}}{u_{n}}}right|=rho }



  • 当ρ<1时级数收敛

  • 当ρ>1时级数发散

  • 当ρ=1时级数可能收敛也可能发散。



例子



收敛


考虑级数


n=1∞nen{displaystyle sum _{n=1}^{infty }{frac {n}{e^{n}}}}sum _{{n=1}}^{infty }{frac  {n}{e^{n}}}

limn→|an+1an|=limn→|n+1en+1nen|=limn→|n+1en+1⋅enn|=limn→|n+1n⋅enen⋅e|=limn→|(1+1n)⋅1e|=1⋅1e=1e<1.{displaystyle {begin{aligned}lim _{nto infty }left|{frac {a_{n+1}}{a_{n}}}right|&=lim _{nto infty }left|{frac {frac {n+1}{e^{n+1}}}{frac {n}{e^{n}}}}right|\&=lim _{nto infty }left|{frac {n+1}{e^{n+1}}}cdot {frac {e^{n}}{n}}right|\&=lim _{nto infty }left|{frac {n+1}{n}}cdot {frac {e^{n}}{e^{n}cdot e}}right|\&=lim _{nto infty }left|left(1+{frac {1}{n}}right)cdot {frac {1}{e}}right|\&=1cdot {frac {1}{e}}={frac {1}{e}}<1.end{aligned}}}{begin{aligned}lim _{{nto infty }}left|{frac  {a_{{n+1}}}{a_{n}}}right|&=lim _{{nto infty }}left|{frac  {{frac  {n+1}{e^{{n+1}}}}}{{frac  {n}{e^{n}}}}}right|\&=lim _{{nto infty }}left|{frac  {n+1}{e^{{n+1}}}}cdot {frac  {e^{n}}{n}}right|\&=lim _{{nto infty }}left|{frac  {n+1}{n}}cdot {frac  {e^{n}}{e^{n}cdot e}}right|\&=lim _{{nto infty }}left|left(1+{frac  {1}{n}}right)cdot {frac  {1}{e}}right|\&=1cdot {frac  {1}{e}}={frac  {1}{e}}<1.end{aligned}}

因此该级数收敛。



发散


考虑级数


n=1∞enn{displaystyle sum _{n=1}^{infty }{frac {e^{n}}{n}}}sum _{{n=1}}^{infty }{frac  {e^{n}}{n}}



























limn→|an+1an|{displaystyle lim _{nrightarrow infty }left|{frac {a_{n+1}}{a_{n}}}right|}lim _{{nrightarrow infty }}left|{frac  {a_{{n+1}}}{a_{n}}}right|
=limn→|en+1n+1enn|{displaystyle lim _{nrightarrow infty }left|{frac {frac {e^{n+1}}{n+1}}{frac {e^{n}}{n}}}right|}lim _{{nrightarrow infty }}left|{frac  {{frac  {e^{{n+1}}}{n+1}}}{{frac  {e^{n}}{n}}}}right|

=limn→|en+1n+1⋅nen|{displaystyle lim _{nrightarrow infty }left|{frac {e^{n+1}}{n+1}}cdot {frac {n}{e^{n}}}right|}lim _{{nrightarrow infty }}left|{frac  {e^{{n+1}}}{n+1}}cdot {frac  {n}{e^{n}}}right|

=limn→|nn+1⋅en⋅een|{displaystyle lim _{nrightarrow infty }left|{frac {n}{n+1}}cdot {frac {e^{n}cdot e}{e^{n}}}right|}lim _{{nrightarrow infty }}left|{frac  {n}{n+1}}cdot {frac  {e^{n}cdot e}{e^{n}}}right|

=limn→|(1−1n+1)⋅e|{displaystyle lim _{nrightarrow infty }left|(1-{frac {1}{n+1}})cdot eright|}lim _{{nrightarrow infty }}left|(1-{frac  {1}{n+1}})cdot eright|

=1⋅e{displaystyle 1cdot e}1cdot e

=e(>1){displaystyle !,e(>1)}!,e(>1)

因此该级数发散。



不能确定


级数


n=1∞1{displaystyle sum _{n=1}^{infty }1}sum _{{n=1}}^{infty }1

发散,但


limn→|11|=1.{displaystyle lim _{nrightarrow infty }left|{frac {1}{1}}right|=1.}lim _{{nrightarrow infty }}left|{frac  {1}{1}}right|=1.

而级数


n=1∞1n2{displaystyle sum _{n=1}^{infty }{frac {1}{n^{2}}}}sum _{{n=1}}^{infty }{frac  {1}{n^{2}}}

收敛,但


limn→|1(n+1)21n2|=1.{displaystyle lim _{nrightarrow infty }left|{frac {frac {1}{(n+1)^{2}}}{frac {1}{n^{2}}}}right|=1.}lim _{{nrightarrow infty }}left|{frac  {{frac  {1}{(n+1)^{2}}}}{{frac  {1}{n^{2}}}}}right|=1.


参见



  • 根值审敛法

  • 比较审敛法




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