I have to fine the following equality
$ \sum^{n-1}_{m=0} \frac{1}{(n-m)p} = neH_n$
where $H_n = 1 + 1/2 + 1/3 + .... 1/n$
My Solution:
$ \sum^{n-1}_{m=0} \frac{1}{(n-m)p} = \frac{1}{n} + \frac{1}{n-1} + \frac{1}{n-2} + ... + \frac{1}{1} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + .. \frac{1}{n} = \sum^{n}_{m=1} \frac{1}{mp} $
$ \frac{1}{p}\sum^{n}_{m=1} \frac{1}{m} = \frac{1}{p}H_n$
How can I proceed further to get $neH_n$
OMG... extremely sorry I missed one subtle note somewhere in the text of the paper.... $p≈1/ne$ (where again e is the Euler's constant)...
Thanks guys for your help Should I delete the question or post this answer as a separate answer since this is not a very complex problem or innovative solution?