How to find if the sequences a n = n π cos ⁡ ( n π ) a n d a n = ( n 2 ) 1 n are convergent or divergent?

Question

How to find if the sequences       a    n    =  n  π  cos  ⁡  (  n  π  )     a  n  d         a    n    =  (      n    2        )                  1        n             are convergent or divergent?

scoseBexgofvc · Accepted Answer

1)       a    n    =  n  π  cos  ⁡  n  πNote that   cos  ⁡  n  π  =  1  ,  w  h  e  n     n     i  s     e  v  e  n  cos  ⁡  n  π  =  −  1  ,  w  h  e  n     n     i  s     o  d  dThus, the first few terms are:   −  π  ,  2  π  ,  −  3  π  ,  4  π  ,  −  5  π  ,  6  π  ,  .  .  .Thus,       a    n    =  n  π  ,  w  h  e  n     n     i  s     e  v  e  n      a    n    =  −  n  π  ,  w  h  e  n     n     i  s     o  d  d      lim          n      →      ∞      ,      n             o      d      d            a    n    =  −  ∞,       lim          n      →      ∞      ,      n             o      d      d            a    n    =  ∞Hence, the       lim          n      →      ∞      ,      n             o      d      d            a    n   does not exist2)       a    n    =      n                  2        n              =  (      n                  1        n                  )    2  Study the pattern for       n                  1        n            Thus, we guess that the sequence converges and the limit is 1, but we have to justify it.      B    n    =      n                  1        n              =  1  +      x    n    ,      x    n    &amp;gt;  0 (to show       lim          n      →      ∞            a    n    =  1  ⇔      lim          n      →      ∞            x    n    =  0)Now,   (  1  +      x    n        )    n    =  n  ,      x    n    &amp;gt;  0  1  +  n      x    n    +            n      (      n      −      1      )        2        x    n    2    +  .  .  .  =  nSo,             n      (      n      −      1      )        2        x    n    2    &amp;lt;  n  ⇒      x    n    &amp;lt;            2              n        −        1              ,  (  n  &amp;gt;  1  )And       lim          n      →      ∞            x    n    &amp;lt;      lim          n      →      ∞                  2              n        −        1              =  0Therefore,       lim          n      →      ∞            a    n    =  1Thus, the limit   n  π  cos  ⁡  n  π does not exist, the sequence divergesAnd limit       n                  2        n              =  1 (converges to 1)

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