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Why MMA delivers this result and how to interpret or reformulate it?

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Limit[k Sum[1/i, {i, 1, k - 1}], k -> 0]

I would expect 0 as it is an empty sum.

MMA 12.1

Edit:

Both expressions below deliver the same results for n>0. By which formula MMA calculates the value -1 for n=0? In case of analytic continuation there should be a formula that I couldn't find. What does this formula look like?

enter image description here

I have answered by own question, see below.

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    $\begingroup$ Analytic continuation of harmonic numbers. Your sum isn't actually empty, but goes from $1$ to $-1$ in the sense of analytic continuation. $\endgroup$
    – Roman
    Commented Jun 25, 2021 at 16:39
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    $\begingroup$ Indeed, k* Sum[1/i, {i, 1, k - 1}] performs $k H_{k-1}$. $\endgroup$
    – user64494
    Commented Jun 25, 2021 at 17:01
  • $\begingroup$ On other hand, Sum[1/i, {i, 1, -1}] equals 0 and HarmonicNumber[-1] is ComplexInfinity and 0*HarmonicNumber[-1] performs Indeterminate. $\endgroup$
    – user64494
    Commented Jun 25, 2021 at 17:11
  • $\begingroup$ @user64494 that's precisely why I mentioned analytic continuation. $\endgroup$
    – Roman
    Commented Jun 25, 2021 at 17:15
  • $\begingroup$ I don't find in the documentation to Sum something similar to Sum[...,{i,1,-1}]. $\endgroup$
    – user64494
    Commented Jun 25, 2021 at 18:31

3 Answers 3

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Why MMA delivers this result?

To see more clearly what's happening, let's do one function per line.

Limit[k Sum[1/i,{i,1,k-1}],k->0] is the same as

f1=Sum[1/i,{i,1,k-1}] $=H_{k-1}$

f2=k f1 $=kH_{k-1}$

f3=Limit[f2,k->0] $=-1$

The basic issue is Mathematica evaluates the Sum first assuming it is not empty. This is indeed an issue you can encounter in Mathematica and can be a bit annoying. The basic issue is Sum will give a general form assuming some things about your variables. You can explicitly see these assumptions by using GenerateConditions. Consider the following:

Sum[1/i,{i,1,-1}] $=0$

Sum[1/i, {i, 1, k-1}] $=H_{k-1}$

Sum[1/i, {i, 1, k-1}]/. k->0 $=$ ComplexInfinity

Sum[1/i,{i,1,k-1},GenerateConditions->True] $=H_{k-1}\text{ if }k\in \mathbb{Z}\land k\geq 2$

Sum[1/i,{i,1,k-1},GenerateConditions->True]/. k->0 $=$ Undefined

and how to interpret or reformulate it?

I've encountered this a few times in the past and never found an elegant solution. What I ended up doing was just defining the special cases separately. For your case that's simple. Just plug in $k=0$ initially:

With[{k = 0}, k Sum[1/i, {i, 1, k - 1}]] gives 0.

In more complicated situations you can use Piecewise, If, or Condition to control the evaluation at a lower level.

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    $\begingroup$ I'm aware, it's the first example I gives under "Consider the following:". I was just showing the more general form - it's good to avoid hard-coding variables when possible. $\endgroup$
    – bRost03
    Commented Jun 25, 2021 at 18:16
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    $\begingroup$ I'm not sure what the relevance of that is? $\endgroup$
    – bRost03
    Commented Jun 25, 2021 at 18:30
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    $\begingroup$ I assume it's not in the documentation because it's fairly common knowledge that the empty sum is $0$ and follows from basic properties of sums. It's empty because $\sum_{i=a}^b$ means $i$ takes on values $a\leq i \leq b$ and there are no values of $i$ such that $1\leq i \leq -1$. $\endgroup$
    – bRost03
    Commented Jun 25, 2021 at 18:39
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    $\begingroup$ I'm not sure how my words were emotional, but I legitimately don't understand the source of your confusion. Please elaborate and I'll give you a detailed argument. Do you not buy that the empty sum should be 0? Or that it's common knowledge that the empty sum is $0$? Or that $\sum_{i=a}^b f(i) \ ; \ b<a$ is empty? Or that Sum[1/i,{i,1,-1}] is empty? Or what? $\endgroup$
    – bRost03
    Commented Jun 25, 2021 at 18:48
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    $\begingroup$ Sorry, your statement " it's fairly common knowledge that the empty sum is 0 and follows from basic properties of sums." is not grounded. I don't find it in the documentation. Can you give a reference? TIA. $\endgroup$
    – user64494
    Commented Jun 25, 2021 at 18:56
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@Roman explains the contradiction by a generic result of Sum. There is a way to avoid any contradiction:

Sum[1/i, {i, 1, k - 1}, GenerateConditions -> True]

ConditionalExpression[HarmonicNumber[-1 + k], Element[k, Integers] && k >= 2]

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  • $\begingroup$ I don't find in the documentation to Sum someting similar to Sum[...,{i,1,-1}]. $\endgroup$
    – user64494
    Commented Jun 25, 2021 at 18:25
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The Harmonic number $H_{n-1}$ can be expressed by the Digamma function $\psi(n)$ together with the Euler-Mascheroni constant $\gamma\approx 0.5772$.

$\psi(n)=H_{n-1}-\gamma$

In MMA $\psi(n)$ and $\gamma$ are called by PolyGamma[n] and EulerGamma.

Table[Limit[(PolyGamma[k] + EulerGamma)*k, k -> n], {n, 0, 4}]
*( {-1, 0, 2, 9/2, 22/3} *)

The limes is only necessary for $n=0$, i.e.

Table[(PolyGamma[n] + EulerGamma)*n, {n, 0, 4}]
(* {Indeterminate, 0, 2, 9/2, 22/3} *)

An expression (inspired from here) that delivers values for all integers $n\ge 0$ without using Limit or control commands is

$H_{n-1}=(n-1) \left(1+n \sum _{i=2}^{\infty } \frac{1}{i (i+n-1)}\right)$

Table[(n-1)(1+n*Sum[1/(i(n-1+i)),{i,2,\[Infinity]}]),{n,0,4}]
*( {-1, 0, 2, 9/2, 22/3} *)
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