I know that the above limit is 1 from Wolfram but I had no proper way to prove. I attempted a proof and it is as follows: lim n → ∞ ( 1 + 1 2 + 1 3 + ⋯ + 1 n log ⁡ n ) = ∫ 0 1 1 n d n lim n → ∞ log ⁡ n .

Question

I know that the above limit is 1 from Wolfram but I had no proper way to prove. I attempted a proof and it is as follows:            lim              n        →        ∞                    (                                    1            +                                          1                2                                      +                                          1                3                                      +            ⋯            +                                          1                n                                                          log            ⁡            n                              )        =                                        ∫            0            1                                              1              n                                d          n                                      lim                          n              →              ∞                                log          ⁡          n                      .

Konner Sloan · Accepted Answer

Your statement            lim              n        →        ∞                    (                                    1            +                                          1                2                                      +                                          1                3                                      +            ⋯            +                                          1                n                                                          log            ⁡            n                              )        =                                        ∫            0            1                                              1              n                                d          n                                      lim                          n              →              ∞                                log          ⁡          n                    makes absolutely no sense. The numerator on the RHS      ∫    0    1        1    n      d  nis a fixed constant. Moreover, your reuse of n as the variable of integration is inappropriate. Finally, the distribution of the limit into the numerator and denominator is also incorrect; e.g.,      lim          x      →      0                  sin      ⁡      x        x    ≠                    lim                  x          →          0                    sin      ⁡      x                      lim                  x          →          0                    x        =      0    0    .Instead, recall that for all   x  ≥  1  x  −  1  &amp;lt;  ⌊  x  ⌋  ≤  x  ,so that      1    x    &amp;lt;      1          ⌊      x      ⌋        ≤      1          x      −      1        .Then integrating this expression from   x  =  k to   x  =  k  +  1 yields  log  ⁡            k      +      1        k    =      ∫          x      =      k              k      +      1            1    x      d  x  &amp;lt;      ∫          x      =      k              k      +      1            1          ⌊      x      ⌋          d  x  ≤      ∫          x      =      k              k      +      1            1          x      −      1          d  x  =  log  ⁡      k          k      −      1        .But the middle integral is simply   1      /    k, since for all   x  ∈  [  k  ,  k  +  1  ). Therefore, we have established  log  ⁡  (  k  +  1  )  −  log  ⁡  k  &amp;lt;      1    k    ≤  log  ⁡  k  −  log  ⁡  (  k  −  1  )  ,for positive integers   k  &amp;gt;  1, and for   k  =  1, obviously   1      /    k  ≤  1. Now summing from   k  =  1 to n and observing the telescoping, we get  log  ⁡  n  ≤  log  ⁡  (  n  +  1  )  ≤      ∑          k      =      1        n        1    k    ≤  1  +  log  ⁡  n  .This allows us to obtain a sufficiently tight bound by which the limit may be established through the squeeze theorem:  1  ≤                    ∑                  k          =          1                n                    1        k                    log      ⁡      n        ≤      1          log      ⁡      n        +  1.

I know that the above limit is 1 from Wolfram but I had no proper way to prove. I attempted a proof

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New Questions in Calculus 1