Ln Series SummationI have been given: ln ⁡ n = 1 1 + 1 2 + 1 3 + ⋯ + 1 n for sufficiently large nWhich I can equate to ln ⁡ ( n ) = ∑ i = 1 n 1 i The series I need to sum is: 1 1 + 1 3 + 1 5 + ⋯ + 1 79999 I know this can be determined by doing ln ⁡ ( 80000 ) − 1 2 ln ⁡ ( 40000 ), but is it valid to say: ln ⁡ 2 n − 1 = ∑ i = 1 n 1 2 i − 1 = 1 1 + 1 3 + 1 5 + ⋯ + 1 2 n − 1 ?If not, a brief reason why it isn't valid would be appreciated!Thanks in advancePS. This is part of a question from the Cambridge maths paper STEP III 2008

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Ln Series SummationI have been given:  ln  ⁡      n    =      1    1    +      1    2    +      1    3    +  ⋯  +      1    n     for sufficiently large   nWhich I can equate to   ln  ⁡  (  n  )  =      ∑          i      =      1        n        1    i  The series I need to sum is:      1    1    +      1    3    +      1    5    +  ⋯  +      1    79999  I know this can be determined by doing   ln  ⁡  (  80000  )  −      1    2    ln  ⁡  (  40000  ), but is it valid to say:   ln  ⁡      2    n    −    1    =      ∑          i      =      1        n        1          2      i      −      1        =      1    1    +      1    3    +      1    5    +  ⋯  +      1          2      n      −      1         ?If not, a brief reason why it isn&#039;t valid would be appreciated!Thanks in advancePS. This is part of a question from the Cambridge maths paper STEP III 2008

Cahokiavv · Accepted Answer

∑          i      =      1        n        1          2      i      −      1        =      ∑          i      =      1              2      n            1    i    −      ∑          i      =      1              n            1          2      i        ∼  log  ⁡      (    2    n    )    −      1    2    log  ⁡      n    =  log  ⁡      2    +      1    2    log  ⁡      n  Meanwhile, in the limit as   n  →  ∞  log  ⁡      (    2    n    −    1    )    =  log  ⁡      2    +  log  ⁡            (      n      −              1        2            )        =  log  ⁡      2    +  log  ⁡      n    +  log  ⁡            (      1      −              1                  2          n                    )        ∼  log  ⁡      2    +  log  ⁡      n  These approximations simply do not match, even to the lowest order. So, no, they are not the same.

Brenden Tran · Accepted Answer

The answer is no.   L  n  (  3  )  =  1  +      1    2    +      1    3   If your formula were valid, it would be   1  +      1    3  Also, the usual name for these sums is the harmonic numbers, denoted       H    n

Ln Series Summation I have been given: ln ⁡ <mrow class="MJX-TeXAtom-OR

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