Using the trigonometric identity of sin ⁡ 2 α = 2 sin ⁡ α cos ⁡ α, I rewrote the expression to: lim n → ∞ ( sin ⁡ ( 2 1 ) n 1 cos ⁡ 1 + ⋯ + sin ⁡ ( 2 n ) n n cos ⁡ n ) = lim n → ∞ ( 2 sin ⁡ ( 1 ) n 1 + ⋯ + 2 sin ⁡ ( n ) n n ) I then tried using the squeeze theorem to get the limit, but I can't get an upper sequence that converges to zero, and the best I could get is a sequence that converges to two: 2 sin ⁡ ( 1 ) n 1 + ⋯ + 2 sin ⁡ ( n ) n n ≤ 2 n 1 + ⋯ + 2 n n ≤ 2 n ⋅ n ⟶ 2What am I missing?

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Using the trigonometric identity of   sin  ⁡  2  α  =  2  sin  ⁡  α  cos  ⁡  α, I rewrote the expression to:      lim          n      →      ∞            (                  sin        ⁡        (        2                  1                )                    n                  1                cos        ⁡                  1                      +    ⋯    +                  sin        ⁡        (        2                  n                )                    n                  n                cos        ⁡                  n                      )    =      lim          n      →      ∞            (                  2        sin        ⁡        (                  1                )                    n                  1                      +    ⋯    +                  2        sin        ⁡        (                  n                )                    n                  n                      )  I then tried using the squeeze theorem to get the limit, but I can&#039;t get an upper sequence that converges to zero, and the best I could get is a sequence that converges to two:            2      sin      ⁡      (              1            )              n              1              +  ⋯  +            2      sin      ⁡      (              n            )              n              n              ≤      2          n              1              +  ⋯  +      2          n              n              ≤      2    n    ⋅  n  ⟶  2What am I missing?

Brendon Fernandez · Accepted Answer

The inequality      ∑          k      =      1              n                  1              k              ≤  1  +      ∫          1              n                  1              x                    d    x  ≤  2      n    ,yields      |                  2        sin        ⁡        (                  1                )                    n                  1                      +    .    .    .    +                  2        sin        ⁡        (                  n                )                    n                  n                      |    ≤  4      1          n        →  0.Note that   sin  ⁡  (  ⋅  ) can change signs, so absolute value is needed in the estimate.

Using the trigonometric identity of sin ⁡ 2 α =

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New Questions in Calculus 1