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