Can the series $1/(i-0) + 1/(i-1) + 1/(i-2) + \cdots$ be reduced to $\log(i)$ Can the series $\dfrac1{i-0} + \dfrac1{i-1} + \dfrac1{i-2} + \cdots$ be reduced to $\log(i)$?
It looked similar to the harmonic series, so I checked wikipedia for Harmonic series, and found the following info. 

So I was just wondering whether it could be simplified to $\log(i)$. Please help me solve it.

$\cfrac{1}{i-n} \sim \cfrac{-1}{n}$
So $\sum \cfrac{1}{i-n}$ converges $\Leftrightarrow$ $\sum \cfrac{-1}{n}$ converges
But as you know, $\sum\limits_{k=1}^n \cfrac{-1}{n}\xrightarrow{n \rightarrow +\infty} - \infty$
$\sum\limits_{k=1}^n \cfrac{1}{i-n}$ doesn't converge.

Another way of doing it would be to use $\cfrac{1}{i-n} = \cfrac{-i-n}{1+n^2}$
$\forall n \ge 1, \Re( \cfrac{1}{i-n} ) = \cfrac{-n}{1+n^2} \le \cfrac{-n}{n+n^2} = \cfrac{-1}{1+n} $
So $\sum\limits_{k=1}^n\Re( \cfrac{1}{i-n} ) \le \sum\limits_{k=1}^n \cfrac{-1}{1+n}\xrightarrow{n \rightarrow +\infty} - \infty$
So $\Re( \sum\limits_{k=1}^n \cfrac{1}{i-n} ) = \sum\limits_{k=1}^n\Re( \cfrac{1}{i-n} )\xrightarrow{n \rightarrow +\infty} - \infty$
So since $\sum\limits_{k=1}^n \cfrac{1}{i-n}$ converges $\Leftrightarrow$ $\Re( \sum\limits_{k=1}^n \cfrac{1}{i-n} )$ converges and $\Im( \sum\limits_{k=1}^n \cfrac{1}{i-n} )$ converges
$\sum\limits_{k=1}^n \cfrac{1}{i-n}$ doesn't converge.
