The topic of this article, the idea of attaching a ﬁnite value to divergent series, is no longer a purely mathematical exercise. These ﬁnite values of divergent series have found application in String theory and quantum ﬁeld theory (Casimir eﬀect).

The ﬁnite sum of real/complex numbers is always ﬁnite. The inﬁnite sum of real/complex numbers can be either ﬁnite or inﬁnite. For instance,

is ﬁnite and

If an inﬁnite sum has ﬁnite value it is said to converge, otherwise it is said to diverge. Divergence do not always mean it grows to $\pm \infty $. For instance,

are diverging, while

is converging. The convergence of a series is deﬁned by the convergence of its partial sum. For instance,

diverges because its partial sum

diverges, as $n\to \infty $. The geometric series

$$\sum _{k=0}^{\infty}{z}^{k}=\frac{1}{1-z},$$ | (1) |

for all $\left|z\right|<1$, because its partial sum

Note that the map $T\left(z\right)=\frac{1}{1-z}$ is well-deﬁned between $T:\u2102\setminus \left\{1\right\}\to \u2102$. In fact, $T$ is a holomorphic (complex diﬀerentiable) functions and, hence, analytic. The analytic function $T$ exists for all complex numbers except $z=1$ and, inside the unit disk $\left\{z\in \u2102\mid \left|z\right|<1\right\}$, $T$ is same as the power series in (1). Thus, $T$ may be seen as an analytic continuation of

outside the unit disk, except at $z=1$. A famous unique continuation result from complex analysis says that the analytic continuation is unique. Thus, one may consider the value of $T\left(z\right)$, outside the unit disk, as an ‘extension’ of the divergent series

In this sense, setting $z=2$ in $T\left(z\right)$, the divergent series may be seen as taking the ﬁnite value

Similarly, setting $z=-1$ in $T\left(z\right)$, the oscillating divergent series may be seen as taking the ﬁnite value

This sum coincides with the Cesàro sum which is a special kind of convergence for series which do not diverge to $\pm \infty $. For instance,

diverges in Cesàro sum too. The Cesàro sum of a series

is deﬁned as the

Observe that the sequence

Since $T$ is not deﬁned on $1$, we have not assigned any ﬁnite value to the divergent series

The Riemann zeta function

converges for $z\in \u2102$ with $\Re \left(z\right)>1$. Note that for $z\in \u2102$, ${k}^{z}={e}^{zln\left(k\right)}$, where $ln$ is the real logarithm. If $z=\sigma +it$ then ${k}^{z}={k}^{\sigma}{e}^{itln\left(k\right)}$ and $\left|{k}^{z}\right|={k}^{\sigma}$ since $\left|{e}^{itln\left(n\right)}\right|=1$. Therefore, the inﬁnite series converges for all $z\in \u2102$ such that $\Re \left(z\right)>1$ (i.e. $\sigma >1$) because it is known that ${\sum}_{k=1}^{\infty}{k}^{-\sigma}$ converges for all $\sigma >1$ and diverges for all $\sigma \le 1$.

The Riemann zeta function is a special case of the Dirichlet series

with ${a}_{k}=1$ for all $k$. Note that for $z=2$, we get the Basel series

and $\zeta \left(3\right)=1.202056903159594\dots $ called the Apéry’s constant. In fact, the following result of Euler gives the value of Riemann zeta function for all even positive integers.

Theorem 1 (Euler). For all $k\in \mathbb{N}$,

where ${B}_{2k}$ is the $2k$-th Bernoulli number.

If $\zeta \left(z\right)$ is well-deﬁned for $\Re \left(z\right)>1$ then ${2}^{1-z}\zeta \left(z\right)$ is also well-deﬁned for $\Re \left(z\right)>1$. Therefore,

Observe that $\eta \left(z\right)$ is also a Dirichlet series. It can be shown that $\left(1-{2}^{1-z}\right)\eta \left(z\right)$ is analytic for all $\Re \left(z\right)>0$ and $\Re \left(z\right)\ne 1$. Thus, we have analytically continued $\zeta \left(z\right)$ for all $\Re \left(z\right)>0$ and $\Re \left(z\right)\ne 1$. There is a little work to be done on the zeroes of $1-{2}^{1-z}$ but is ﬁxable. In the strip $0<\Re \left(z\right)<1$, called the critical strip, the Riemann zeta function satisﬁes the relation

This relation is used to extend the Riemann zeta function to $z$ with non-positive real part, thus, extending to all complex number $z\ne 1$. Setting $z=-1$ in the relation yields

Since $\zeta \left(-1\right)$ is extension of the series

one may think of $\zeta \left(-1\right)$ as the ﬁnite value corresponding to $1+2+3+\dots $. The value of $\zeta \left(0\right)$ is obtained limiting process because it involves $\zeta \left(1\right)$ which is not deﬁned. However, there is an equivalent formulation of Riemann zeta function for all $z\in \u2102\setminus \left\{1\right\}$ as

with a simple pole and residue $1$ at $z=1$.

Recall that the analytic continuation $T$ of the geometric series is not deﬁned for $z=1$ and, hence, we could not assign a ﬁnite value to

But, setting $z=0$ above,

where ${\delta}_{0k}$ is the Kronecker delta deﬁned as

The harmonic series

corresponds to $\zeta \left(1\right)$ which is not deﬁned. The closest one can conclude about $\zeta \left(1\right)$ is that

where $\gamma $ is the Euler-Mascheroni constant which has the value $\gamma =0.57721566\dots $.