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I know the sum of the reciprocals of the natural numbers diverges to infinity, but I want to know what value can be assigned to it.

$$\sum_{n=1}^{\infty}\frac1n=\frac11+\frac12+\frac13+\frac14+\dots=L$$

As a few examples of what kind of answer I want, here are a few similar problems:

$$\sum_{n=1}^{\infty}n=1+2+3+4+\dots=-1/12$$

$$\sum_{n=1}^{\infty}(-1)^{n+1}n=1-2+3-4+\dots=1/4$$

$$\sum_{n=0}^{\infty}(-1)^n=1-1+1-1+\dots=1/2$$

As you can see, I want to assign a value to the divergent series of the reciprocals of the natural numbers.

Simply Beautiful Art
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    Lucian's reference to $\gamma$ as a potential result has support (links provided in a comment at the end of his answer). – Brian Tung Jan 30 '16 at 00:24
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    Mentioned in some answer comment, deserves to be here http://mathoverflow.net/questions/3204/does-any-method-of-summing-divergent-series-work-on-the-harmonic-series – leonbloy Jan 31 '16 at 02:39
  • For anyone who cares: https://en.wikipedia.org/wiki/Ramanujan_summation#Sum_of_divergent_series – Simply Beautiful Art Feb 01 '16 at 23:17
  • @reuns Are you sure that regulator is correct? Seems to be diverging for me. – user76284 Jun 28 '19 at 04:37
  • @user76284 Then look at $$\lim_{t \to 0} \sum_{n=1}^\infty \frac{n^{-1- t(t^3+i)}+n^{-1- t(t^3-i)}}{2} =\lim_{t \to 0} \frac{\zeta(1+ti+t^4)+\zeta(1-ti+t^4)}{2}= \lim_{t \to 0} \frac{\frac{1}{ti+t^4}+\frac{1}{-ti+t^4}+2\gamma + O(t)}{2}=\lim_{t \to 0} \frac{\frac{1}{ti}+O(t^3)+\frac{1}{-ti}+O(t^3)+2\gamma + O(t)}{2}=\gamma$$ And replace $t^4$ by $At^2$ to obtain $A+\gamma$ as the limit – reuns Jun 28 '19 at 14:09
  • @reuns Apologies if this is a stupid question but when I substitute a large $N$ for infinity I get the following: https://imgur.com/a/wRSybxg. So it doesn't seem to actually be converging? – user76284 Jun 28 '19 at 17:29
  • @user76284 You need to let $N \to \infty$ faster than $t \to 0$, keeping $N$ fixed and letting $t\to 0$ you'll get $\sum_{n=1}^N \frac1n$. But anyway you'll never get a good numerical approximation as you need to take $t = 0.001$ to obtain $ \frac{\zeta(1+ti+t^4)+\zeta(1-ti+t^4)}{2} \approx 0.578$ and to obtain that precision you'll need something like $N = 10^{12}$ – reuns Jun 28 '19 at 17:45
  • @reuns Can you explain what you mean by letting $N \to \infty$ faster than $t \to 0$? Increasing $N$ seems to blow things up near 0, regardless of fhe value of $t$ in that region. – user76284 Jun 29 '19 at 07:21
  • For $t = 10^{-3}$ then $t^4 = 10^{-12}$ so you need $N$ very large to obtain $\zeta(1+ti+t^4) \approx \sum_{n=1}^N n^{-1-it-t^4}$ – reuns Jun 29 '19 at 14:54
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    Does this answer your question? Is it possible to use regularization methods on the Harmonic Series? – tparker Dec 04 '21 at 17:19

5 Answers5

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Can we assign a value to the sum of the reciprocals of the natural numbers ?


To quote Obama: Yes, we can ! $~$ Two, in fact. One such possible value would be the

Euler-Mascheroni constant, since $~\dfrac{\zeta\big(1^+\big)+\zeta\big(1^-\big)}2~=~\gamma~.~$ Another one would be $\ln2,$

since in many formulas1 where one would symbolically expect $\zeta(1)$ to be present, $\ln2$

appears there instead.

1 Since many have inquired about this particular statement: There are many parametric

infinite series $S_n,$ as well as many parametric definite integrals $I_n,$ for which the general

formula is a linear combination of terms of the form $\zeta(k)\cdot\zeta(n-k),~$ $\zeta(n-mk)\cdot\zeta(k)^m,$

and $\zeta(n-k)\cdot\ln^k2.$

Lucian
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  • whoa whoa, hold on! How did you arrive to this conclusion? Please explain. – Simply Beautiful Art Jan 26 '16 at 21:43
  • @SimpleArt: Which part do you have difficulties with ? – Lucian Jan 26 '16 at 21:55
  • How did you derive that answer? To start, how is it equivalent to the Euler-Mascheroni constant? I don't understand how you got there. – Simply Beautiful Art Jan 26 '16 at 21:57
  • @SimpleArt: It is mentioned in the link I posted. – Lucian Jan 26 '16 at 22:02
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    It is not enough to supply a link and assume a reader can deduce what you have deduced, nor is it an acceptable answer to my question. Please provide more insight. – Simply Beautiful Art Jan 26 '16 at 22:04
  • @SimpleArt: I didn't deduce it, so I'm afraid I can't provide any insight, since I myself don't possess any. – Lucian Jan 26 '16 at 22:10
  • Aw, that truly is disappointing. Could you at least point to where you see this result appearing in your link? – Simply Beautiful Art Jan 26 '16 at 22:11
  • @SimpleArt: In the middle of the section entitled Relation to the zeta function. – Lucian Jan 26 '16 at 22:13
  • Well, first I note that it is a limit that may at first sight seem to be the sum of the reciprocal of the harmonic numbers, but I don't think it is. Secondly, it is a limit, and limits tend not to exist when dealing with diverging summations such as mine. – Simply Beautiful Art Jan 26 '16 at 22:16
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    Possibly relevant to this line of reasoning: http://math.stackexchange.com/questions/20005/is-it-possible-to-use-regularization-methods-on-the-harmonic-series and http://mathoverflow.net/questions/3204/does-any-method-of-summing-divergent-series-work-on-the-harmonic-series – Brian Tung Jan 30 '16 at 00:23
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    The wikipedia article https://en.wikipedia.org/wiki/Riemann_zeta_function says the Cauchy principal value (the average of the limit from above and the limit from below) is the Euler-Mascheroni constant in 1. I don't have any addition insight, but that seems like the logical choice. – Alon Navon Jan 30 '16 at 20:16
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    Would tending to 1 from four different directions in the complex plane lead to $\frac{\gamma}{2}$? – Jaume Oliver Lafont Jan 30 '16 at 21:37
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    when $|\epsilon| \to 0$ : $\ \ \zeta(1+\epsilon) = 1/\epsilon + \gamma + \mathcal{O}(|\epsilon|)$ so the direction from which it tends to $1$ doesn't matter, as far as the $1/\epsilon$ divergent term cancels out – reuns Jan 30 '16 at 22:16
  • @Lucian: do you have an information/a link where the $\ln(2)$-relation is explained? (I've arrived at such a relation but my method is experimental so I'd like to backup or correct my arguing and possibly get deeper insight) – Gottfried Helms Feb 01 '16 at 09:35
  • @GottfriedHelms: In the past, on this very site, I've come across many series and integrals, whose general formula suggests such a relation, but that was $1$ or $2$ years ago, and I can't locate them. – Lucian Feb 01 '16 at 11:10
  • @GottfriedHelms: a formula where both $log(2)$ and $\gamma$ appear is http://www.wolframalpha.com/input/?i=lim+n-%3Einf+(log(2n%2B1)-HarmonicNumber%5Bn%5D) Incidentally, note that $log(2)-\gamma$ < $\gamma$, so $log(2n+1)$ is closer to $H_n$ than the usual $log(n)$ or $log(n+1)$ – Jaume Oliver Lafont Feb 01 '16 at 11:56
  • @Jaume : wow, that's a curious formula. But like Lucian I thought I'd seen a direct relation of the $\ln(2)$ -substitution anywhere (but without concrete idea where - I think I should reconsider Konrad Knopp's monography on series). I'm experimenting with the matrix of Eulerian numbers and created a summation-procedure from this (partially described in a question at http://math.stackexchange.com/questions/310123). It works well for convergent and many divergent series (even for the infamous 0!-1!+2!-3!+...) and arrives for the nonalternating harmonic series at the $\ln(2)$ – Gottfried Helms Feb 01 '16 at 13:13
  • @GottfriedHelms: It is quite possible that the $\zeta$ there should be an $\eta,$ especially since $\eta(1)=\ln2,$ and the relation between the Riemann $\zeta$ and Dirichlet $\eta$ functions is well-known and well-researched. Also, some of the results I mentioned are to be found among Cleo's $40$ answers, or by googling the site for expressions like "Euler sum", "Euler sums", or "Euler series". – Lucian Feb 01 '16 at 13:45
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    @GottfriedHelms A manipulation that transforms the harmonic series into log(2) is given by $$a(n)=\frac{1}{n}, n>0$$ $$b(n)=a(n)$$ if $n$ is odd $$b(n)=a(n)-a(\frac{n}{2})$$ if $n$ is even Now $\sum_{n=1}^\infty b(n)=log(2)$, but there is nothing magical about 2 here, this procedure gives log(n) when generalizing. – Jaume Oliver Lafont Feb 01 '16 at 14:03
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    @Jaume , (at) Lucian: I just had a look in my working paper about the "Eulerian summation" on nonalternating divergent series and the $\ln(2)$ - bit was not the "result" but rather somehow a residue. If you like to read into this it is in http://go.helms-net.de/math/binomial_new/EulerianSumsV2.pdf at page 15 and 16. (Unfortunately I don't have much mathematical conversation so the writing style is (as said by some correspondent) at many places still difficult to read - but perhaps interesting enough to give it a try anyway...) – Gottfried Helms Feb 01 '16 at 14:23
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    @GottfriedHelms Another way to get almost $log(2)$ is evaluating the z-transform of sequence $$\frac{1}{n+1}, n>=0$$ at the following available integer (z=2), given that for z=1 it does not converge: $$\sum_{n=0}^\infty\frac{1}{(n+1)2^n}=2log(2)$$ – Jaume Oliver Lafont Feb 01 '16 at 15:53
  • This is true. There are tons of times where i'd expect to find something about \zeta(1) even some alternating series.which gives a sorta convergent value, and i'm stuck with this ln(2). Never been been able to lay my finger at the cause of this because the constant appears just as often. – Gerben Mar 19 '16 at 03:13
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Two families of series include the harmonic series and constants similar to the ones given by @Lucian, $\gamma$ and $log\left(2\right)$.

The one related to $\gamma$ is

$$ \gamma= \lim_{n \to \infty} {\left(2H_n-H_{n^2} \right)}=\sum_{n=1}^\infty \left(\frac{2}{n}-\sum_{j=(n-1)^2+1}^{n^2} \frac{1}{j}\right) $$

(for more variants see Series for Stieltjes constants from $\gamma= \sum_{n=1}^\infty \left(\frac{2}{n}-\sum_{j=n(n-1)+1}^{n(n+1)} \frac{1}{j}\right)$)

and the one related to $log\left(2\right)$ is

$$ \log\left(\frac{p}{q}\right)= \lim_{n \to \infty} {\left(H_{pn}-H_{qn}\right)}=\sum_{i=0}^\infty \left(\sum_{j=pi+1}^{p(i+1)}\frac{1}{j}-\sum_{k=qi+1}^{q(i+1)}\frac{1}{k}\right) $$

(https://math.stackexchange.com/a/1593145/134791)

Paying attention to the positive terms only, the first series suggests a link between the harmonic series and $\frac{\gamma}{2}$, while in the second one the harmonic series may be related to the logarithm of any positive rational, $log\left(\frac{p}{q}\right)$.

[EDIT] There is an important difference between the three examples mentioned and the one in the question. If we define the cancelling series of a series as the following operation: $$ s_{0}=\sum_{k=0}^\infty\left(a(k)-a(k)\right) $$

then the cancelling series of the three examples provided is still a divergent series, while the cancelling series of the harmonic series becomes conditionally convergent. In other words, methods for divergent series are not needed (and need not work) since this cancelling method lets assign a value for the harmonic series (and does not "work" for other divergent series). This value obtained is 0 in principle, as in the answer by @NickS. This is a direct consequence of the definition of cancelling series, which seems trivial, but it is interesting that not all divergent series have a convergent cancelling series. However, since the cancelling series is conditionally convergent, this zero becomes a particular case of $log\left(\frac{p}{q}\right)$ under Riemann rearrangements. Another particular case is $log(2)$ for $p=2q$.

Community
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Jaume Oliver Lafont
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We should not write any of those as "equations" without explanation. If you use a summation method, say what it is: in the text, or even put it somewhere on your $\sum$ notation.

As noted, different summation methods (for a divergent series) may give you different answers. And different summation methods may be useful for different purposes; or---in many cases---for no known purpose.

GEdgar
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    I'd disagree. If you integrate you can also find the indefinite intragal, heck that's the first thing you learn. Why be so rigid, a normal summation will only be part of the solution. If it diverge the next intresting thing is what the constant term will be. It's like being an explorer, sometime you got to take risk to find new thing. This image of the rigid mathematician is the reason i'm studying economics, the area of the brave and free. – Gerben Mar 19 '16 at 03:19
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Here is a thought. We can write the harmonic series as differences, so $1+\frac{1}{2}+\frac{1}{3}+...$=$2-1+\frac{2}{2}-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}...$. What we can note is that we have a negative harmonic sequence. So, for any positive number in the series, we can estimate it arbitrarily closely by negative members of the harmonic sequence as said sequence goes to 0. Also, we will not run out of estimation material since the harmonic series diverges from any starting point. Thus, we can write this whole mess as $\sum \epsilon_1-\epsilon_2+\epsilon_3-\epsilon_4...$ where the epsilons are close to anything we please. So I would conjecture that we can assign a "0" to the harmonic series. Of course, now that we have done this, by the Riemann Rearrangement theorem, we can assign any value we please.

Nick S
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  • We can clearly assign a 0 to the cancelling harmonic series (the sum of the positive and negative harmonic series at the same pace http://math.stackexchange.com/a/1602987/134791 or log(1) in my answer to this question) – Jaume Oliver Lafont Jan 30 '16 at 21:10
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We can define the sum of the Harmonic series by $$ \sum_{n=1}^\infty \frac{1}{n} = \frac{\eta'(1)}{\log(2)} = \gamma - \frac{\log(2)}{2}. $$

A series $\sum_{n=1}^\infty b_n$ is Abel summable to $g(1)$ if (i) $g(x) = \sum_{n=1}^\infty b_n x^n$ converges for $|x| < 1$ (at least) and (ii) there exists the limit $g(1) = \lim_{x \to 1^-} g(x)$.

A series $\sum_{n=1}^{\infty} a_n$ belongs to the Ramanujan class $[R]$ if $f(x) = \sum_{n=1}^\infty a_n x^n$ and $f(x) - R f(x^2) = g(x)$, or rather, (i) $a_{2k-1} = b_{2k-1}$, $a_{2k} = b_{2k} + R a_k$ (so that $a_{2k}$ is a polynomial in $R$) and (ii) $\sum_{n=1}^\infty b_n$ is Abel summable.

For $R \neq 1$, $f(1) = \sum_{n=1}^\infty a_n$ is elementary Ramanujan summable to $$ \frac{g(1)}{1 - R}. $$ This summation is a linear function defined in $[R]$ (which is a linear space). It agrees with the analytic continuation of the Dirichlet series $F(s) = \sum_{n=1}^\infty a_n n^{-s}$ at $s=0$. In fact, for $G(s) = \sum_{n=1}^\infty b_n n^{-s}$, the recursive relations between the coefficients $a_n$ and $b_n$ imply that $(1 - R 2^{-s}) F(s) = G(s)$. Thus, $$ F(s) = \frac{G(s)}{1 - R 2^{-s}} $$ and $$ F(0) = \frac{G(0)}{1 - R}. $$ For $g(1) = G(0)$, $f(1) = F(0)$.

A summation to be defined in $[1]$ must be a linear function. If $g(1) = G(0) = 0$, the sum of $f(1) = F(0)$ is given by l'Hopital's rule, $$ \sum_{n=1}^\infty a_n = \frac{G'(0)}{\log(2)}. $$

For example, $f(x) = x$, $g(x) = x - x^2$, $F(s) = 1$, $G(s) = 1 - 2^{-s}$, $G'(0) = \log(2)$, $g(1) = G(0) = 0$, $f(1) = F(0) = 1$.

As the derivative $G'(0)$ is linear, the definition is natural in the entire linear space $[1]$.

The Harmonic series corresponds to $f(x) = - \log(1 - x) = \sum_{n=1}^\infty n^{-1} x^n$, $g(x) = \sum_{n=1}^\infty (-1)^{n-1} n^{-1} x^n$, $F(s) = \zeta(s+1) = \sum_{n=1}^\infty n^{-s - 1}$, $G(s) = \eta(s + 1) = \sum_{n=1}^\infty (-1)^{n-1} n^{-s - 1}$. Thus, $R = 1$ and $$ f(1) = F(0) = \frac{\eta'(1)}{\log(2)} = \gamma - \frac{\log(2)}{2}. $$

In general, $$ \sum_{n=1}^\infty a_n = \frac{G'(0)}{\log(2)} + G(0)\sum_{n=1}^\infty \text{pow2}(n), $$ where $\text{pow2}(n)$ is $1$ if $n$ is a power of $2$, $0$ otherwise, and the sum of $\sum_{n=1}^\infty \text{pow2}(n)$ is set by definition. If $\sum_{n=1}^\infty \text{pow2}(n) = 1/2$, then $\sum_{n=1}^\infty 1/n = \gamma$.

This definition is natural, because $$ P(s) = \sum_{n=1}^\infty \text{pow2}(n) n^{-s} = \frac{1}{1 - 2^{-s}} $$ and $$ \frac{P(s) + P(-s)}{2} = \frac{1}{2}. $$

efferrari
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