How can I find the limit of u n = sin ⁡ ( 1 n + 1 ) + ⋯ + sin ⁡ ( 1 2 n )

Question

How can I find the limit of      u    n    =  sin  ⁡      (          1              n        +        1              )    +  ⋯  +  sin  ⁡      (          1              2        n              )

Jovan Wong · Accepted Answer

We can make use of the fact that  sin  ⁡  x  =      ∑          n      =      0        ∞    (  −  1      )    n              x              2        n        +        1                    (      2      n      +      1      )      !      and note that when   x  ≤  1, the absolute value of each term is more than the sum of the absolute values of the later terms. Thus   x  −            x      3        6    &amp;lt;  sin  ⁡  x  &amp;lt;  x, so we have      ∑          k      =      n      +      1              2      n            1    k    −  n      1          6              n        3              &amp;lt;      ∑          k      =      n      +      1              2      n            1    k    −      1          6              k        3              &amp;lt;      u    n    &amp;lt;      ∑          k      =      n      +      1              2      n            1    k  and                              lim                      n            →            ∞                                    ∑                      k            =            n            +            1                                2            n                                    1          k                                    =                              lim                      n            →            ∞                                    (                      ∑                          k              =              1                                      2              n                                            1            k                    −          ln          ⁡          (          2          n          )          )                −                  lim                      n            →            ∞                                    (                      ∑                          k              =              1                                      n                                            1            k                    −          ln          ⁡          (          n          )          )                +                  lim                      n            →            ∞                          (        ln        ⁡        (        2        n        )        −        ln        ⁡        (        n        )        )                                          =                              lim                      n            →            ∞                          (        ln        ⁡        (        2        n        )        −        ln        ⁡        (        n        )        )                                          =                              lim                      n            →            ∞                          ln        ⁡        2                                          =                    ln        ⁡        2            while clearly       lim          n      →      ∞        n      1          6              n        3              =  0, so by squeeze theorem       lim          n      →      ∞            u    n    =  ln  ⁡  2

How can I find the limit of u n </msub> = sin ⁡ (

Answered question

Answer & Explanation

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