# 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.

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The natural logarithm is multivalued on $\mathbb{C}$ mathworld.wolfram.com/NaturalLogarithm.html – xavierm02 Nov 17 '12 at 11:28
Did you intend $i$ to be a real number or $\sqrt{-1}$? – robjohn Nov 18 '12 at 17:24
No, i is just an integer or a variable – Vishnu Vivek Nov 18 '12 at 17:38
If $i$ is intended to be an integer, does the $\dots$ stop before $\frac1{i-n}=\frac10$? That is, is the sum actually finite? – robjohn Nov 18 '12 at 17:55

The real part will not converge because of the $\frac 1n$ equivalence for large $n$ (as shown by others) but the imaginary part admits following limit : $$\tag{1} \sum_{n=0}^\infty \Im\left(\frac 1{i-n}\right)=-\frac {1+\pi\coth(\pi)}2=-\sum_{n=0}^\infty \frac 1{1+n^2}$$

since $\ \displaystyle\Im\left(\frac 1{i-n}\right)=\frac 1{2i}\left(\frac 1{i-n}-\frac 1{-i-n}\right)=-\frac 1{1+n^2}\$

and using the classical series (see for example here) for $\cot(\pi z)$ : $$\tag{2}\pi\ \cot(\pi z)=\frac 1z +\sum_{n=1}^\infty \frac {2z}{z^2-n^2}$$

that we will rewrite as (setting $\ z:=i\,s\$ and using $\,\coth(s)=i\,\cot(i\,s)$) : $$\tag{3}\pi\ \coth(\pi\, s)=i(\pi\,\cot(\pi is))=i\left(-\frac is +\sum_{n=1}^\infty \frac {2is}{-s^2-n^2}\right)=\frac 1s +\sum_{n=1}^\infty \frac {2s}{s^2+n^2}$$

Setting $s:=1$ and adding the $n=0$ term should give us $(1)$.

Concerning the finite sum you considered : $$\sum_{n=1}^k \frac 1n >\int_1^{k+1}\frac {dx}{\ln(x)}=\ln(k+1)$$ it provides in fact a proof that the harmonic sum grows without bounds (remember that you want the limit as $k\to \infty$ i.e. $+\infty$ here).
You may modify this (for example using $\left|\frac 1{i-n}\right|>\frac 1n$) to prove that your sum is divergent but won't get a bounded value like $\ln(i)$.

Hoping this clarified things,

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No. A finite series of partial sums reduces to the special digamma function. It is not possible to express this function in terms of elementary functions

$$\sum_{x=0}^n{\frac{1}{a-x}}=-\sum_{x=0}^n{\frac{1}{x-a}}=~ψ(-a) - ψ(n-a+1)$$

An infinite series of this form does not converge (comaprison with standard harmonic series). In general, sums of this sort aren't simply related to integrals, though there are theorems which could help you express the specified sum using integrals. Namely, the Euler-MacLaurin formula.

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$\cfrac{1}{i-n} \sim \cfrac{-1}{n}$

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.

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It's just a partial answer, I think the answer is no, it should be considered in som other way..

We have $\displaystyle\frac1{1-x}=-\sum_nx^n\$ (if $|x|<1$), hence $\log(1-x)=\displaystyle\sum_{n\ge 1}\frac{x^n}n$.

Plugging in $x=1$, we arrive to $\displaystyle\sum_{n\ge 1}\frac1n=\log 0=(\pm)\infty$, and it seems that this is what was used in what you wanted to generalize.

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