5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Lucian's reference to $\gamma$ as a potential result has support (links provided in a comment at the end of his answer). $\endgroup$ – Brian Tung Jan 30 '16 at 0:24
  • 1
    $\begingroup$ Mentioned in some answer comment, deserves to be here mathoverflow.net/questions/3204/… $\endgroup$ – leonbloy Jan 31 '16 at 2:39
  • $\begingroup$ For anyone who cares: en.wikipedia.org/wiki/… $\endgroup$ – Simply Beautiful Art Feb 1 '16 at 23:17
  • $\begingroup$ @reuns Are you sure that regulator is correct? Seems to be diverging for me. $\endgroup$ – user76284 Jun 28 at 4:37
  • $\begingroup$ @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 $\endgroup$ – reuns Jun 28 at 14:09
9
+100
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Simply Beautiful Art Jan 26 '16 at 22:04
  • 2
    $\begingroup$ Possibly relevant to this line of reasoning: math.stackexchange.com/questions/20005/… and mathoverflow.net/questions/3204/… $\endgroup$ – Brian Tung Jan 30 '16 at 0:23
  • 1
    $\begingroup$ The wikipedia article 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. $\endgroup$ – Alon Navon Jan 30 '16 at 20:16
  • 1
    $\begingroup$ Would tending to 1 from four different directions in the complex plane lead to $\frac{\gamma}{2}$? $\endgroup$ – Jaume Oliver Lafont Jan 30 '16 at 21:37
  • 2
    $\begingroup$ 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 $\endgroup$ – reuns Jan 30 '16 at 22:16
4
$\begingroup$

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 $p=2q$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – Gerben Mar 19 '16 at 3:19
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $$ \frac{P(s) + P(-s)}{2} = \frac{1}{2}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.