I've met the following. Set b n := ∫ 0 π / 2 cos 2 n + 1 ⁡ ( x ) d x .

Question

I&#039;ve met the following. Set      b    n    :=      ∫    0          π              /            2            cos          2      n      +      1        ⁡  (  x  )      d    x  .

podesect · Accepted Answer

First your recursion should yield      ∏          j      =      1              n            (                            2          j                          2          j          +          1                      )  Now note that      b    n    =      ∏          j      =      1              n            (                            2          j                          2          j          +          1                      )    =      ∏          j      =      1              n            (                  j                  j          +          1                      /                    2                      )    =                              ∏                      j            =            1                                n                          j                              ∏                      j            =            1                                n                                    (          j          +          1                      /                    2          )                      =                    π            2                          Γ        (        n        +        1        )                    Γ        (        3                  /                2        )                              ∏                          j              =              1                                      n                                            (            j            +            1                          /                        2            )                              Recall that  Γ  (  z  +  1  )  =  z  Γ  (  z  )           and           Γ  (  3      /    2  )  =      π        /    2  =                    π            2                          Γ        (        n        +        1        )                    Γ        (        n        +        3                  /                2        )            Once you have this you should be able to get the limit by looking at the asymptotics of   Γ  (  z  ). You could either do by say, Stirling&#039;s formula (or) make use of the fact that for large enough z                    Γ        (        z        +        α        )                    Γ        (        z        )              ∼      z          α      where   α  ∈      R

I've met the following. Set b n </msub> := <msubsup> ∫ 0

Answered question

Answer & Explanation

New Questions in Calculus 2