As we can see on Wikipedia, there are some algebraic methods that give us finite sums for the Grandi's series


Let $S$ be the sum of the Grandi's series. Then

  • $S=(1-1)+(1-1)+\cdots=0+0+\cdots=0$

  • $S=1+(-1+1)+(-1+1)+\cdots=1+0+0+\cdots=1$

  • $1-S=S$ so that $S=1/2$

The algebraic manipulations above are not allowed because the Grandi's series doesn't converges.

Is there something similar related to the harmonic series? In other words, is there any "incorrect algebraic way" to get a finite sum for

$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots \;?$$


  • $\begingroup$ Hint: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots=2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+ \cdots)$ $\endgroup$ – Martigan Oct 20 '14 at 9:26
  • $\begingroup$ A related question. $\endgroup$ – Lucian Oct 20 '14 at 10:01
  • 1
    $\begingroup$ @Martigan - Please more guide. $\endgroup$ – H.S Jan 14 '17 at 11:14

Actually, if you only consider the normal harmonic series, with only positive terms, you can't find a rearrangement that will converge to a certain value since all terms are positive, and changing the order is not possible.

However you can prove that, if you re-arrange correctly the terms of the series of general term: $a_n = \frac{(-1)^{n-1}}{n}$ , you can make it converge towards any real x.

Roughly: $a_1 = 1$ , if x<1, add negative terms until $x> S_{f(n)} $ where this partial sum is made with odd terms of the series. Then once you have this condition, you will add positive terms until : x < $S_{f(n)} $ . You do that undefinitely, and putting it properly enough you can prove that this series converges towards x. Of course this in only sketches, but the idea is here

  • $\begingroup$ This is possible because the harmonic series is divergent, and at each step you only add terms of the series of same signs so you are sure to go past x at a certain point. That's interesting. I can maybe provide you with an exam that treat this thouroughly but it will be in french, I'm not sure it will be useful to you $\endgroup$ – mvggz Oct 20 '14 at 10:08

Let $~\zeta^\star(x)=\displaystyle\lim_{h\to0}\dfrac{\zeta(x+h)+\zeta(x-h)}2.~$ Then $\zeta^\star(1)=\gamma\simeq\dfrac1{\sqrt3}.~$ See the Riemann $\zeta$ function

and the Euler-Mascheroni constant for more details. Another possible alternative would be $\ln2$, since it appears in many expressions as a substitute for where one might expect to encounter $\zeta(1)$.



If you know this inequality, it may be helpful for you:


So, $1+\frac12+\frac13+\cdots+\frac1n\le 1+\ln2+\ln\frac32+\cdots \ln\frac{n}{n-1}=1+\ln n$


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