Consider f a bounded, measurable function defined on [ 1 , ∞ ] and define a n := ∫ [ n , n + 1 ) f. Does the fact that f is integrable imply ∑ n = 1 ∞ a n converges?I would like to say that the argument ∑ n = 1 ∞ a n = ∫ 1 ∞ fmade in that post is not correct.My attempt was the following: Since f is integrable, | f | is also integrable. Therefore, ∫ [ 1 , ∞ ) | f | = lim k → ∞ ∫ 1 k | f | &lt; ∞ .Moreover, we have the following: ∑ n = 1 ∞ a n = ∑ n = 1 ∞ ∫ n n + 1 f ≤ ∑ n = 1 ∞ | ∫ n n + 1 f | ≤ ∑ n = 1 ∞ ∫ n n + 1 | f | = lim k → ∞ ∑ n = 1 k ∫ n n + 1 | f | However, I cannot proceed further from here. Can anyone help?

Question

Consider   f a bounded, measurable function defined on   [  1  ,  ∞  ] and define       a    n    :=      ∫          [      n      ,      n      +      1      )        f. Does the fact that   f is integrable imply       ∑          n      =      1        ∞        a    n   converges?I would like to say that the argument      ∑          n      =      1        ∞        a    n    =      ∫    1    ∞    fmade in that post is not correct.My attempt was the following: Since   f is integrable,       |    f      |   is also integrable. Therefore,      ∫          [      1      ,      ∞      )            |    f      |    =      lim          k      →      ∞            ∫    1    k        |    f      |    &amp;lt;  ∞  .Moreover, we have the following:                              ∑                      n            =            1                    ∞                          a          n                                    =                  ∑                      n            =            1                    ∞                          ∫          n                      n            +            1                          f                                          ≤                  ∑                      n            =            1                    ∞                          |                          ∫          n                      n            +            1                          f                  |                                                  ≤                  ∑                      n            =            1                    ∞                          ∫          n                      n            +            1                                    |                f                  |                                                  =                  lim                      k            →            ∞                                    ∑                      n            =            1                    k                          ∫          n                      n            +            1                                    |                f                  |                    However, I cannot proceed further from here. Can anyone help?

Braylon Perez · Accepted Answer

We have                              ∑                      n            =            1                                ∞                                    |                          a          n                          |                                    ≤                  ∑                      n            =            1                                ∞                                    ∫                      [            n            ,            n            +            1            )                                    |                f                  |                                                  =                  ∑                      n            =            1                                ∞                          ∫                  |                f                  |                          1                      [            n            ,            n            +            1            )                                                            =        ∫                  ∑                      n            =            1                                ∞                                    |                f                  |                          1                      [            n            ,            n            +            1            )                                                            =        ∫                  |                f                  |                          1                      [            1            ,            ∞            )                                                            &amp;lt;        ∞        .            The second equality is by MCT and additivity of the integral on nonnegative functions. Alternatively, you can argue directly using DCT that       ∑          n      =      1              ∞            a    n    =  ∫  f      1          [      1      ,      ∞      )      .

Damon Stokes · Accepted Answer

Since   f is integrable, you know that      ∫    1          k      +      1            |    f      |      d  x  =      ∑          n      =      1              k            ∫    n          n      +      1            |    f      |      d  xconverges to       ∫    1    ∞        |    f      |      d  x  &amp;lt;  ∞ as   k  →  ∞ (by monotonic convergence, extending   f by zero beyond   k  +  1). Thus your last limit is finite and your series   ∑      a    n   is convergent

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