Can we sum a series over noninteger bounds?

Question

Are there any ways to generalize series to noninteger bounds? For example, is there any way to make sense of $$ f(\nu)=\sum_{k=\nu}^\infty a_k $$ for $\nu\in\Bbb C$? I would expect that if such a generalization exists, it should obey have the same properties as the case where $\nu$ is an integer. In particular, I would expect it to obey the recurrence relation $$ f(\nu)-f(\nu+1)=a_\nu. $$

Answer 1

I guess you could write something like $$ \sum_{n = 0}^\infty f(e + n) $$

Answer 2

We find in chapter 2 Sums of Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik the following statement (see formula 2.4):

Formally, we write \begin{align*} \sum_{P(k)} a_k\tag{2.4} \end{align*} as an abbreviation for the sum of all terms $a_k$ such that $k$ is an integer satisfying a given property $P(k)$. (A property $P(k)$ is any statement about $k$ that can be either true or false.)

We see the index $k$ is an integer. Sometimes we have the situation that the upper limit of a sum is not an integer. In this situation we take the greatest integer less or equal to the upper limit (floor function). \begin{align*} \sum_{k=0}^{\frac{n}{2}} a_k=\sum_{k=0}^{\color{blue}{\left\lfloor\frac{n}{2}\right\rfloor}} a_k=a_0+a_1+\cdots+a_{\left\lfloor\frac{n}{2}\right\rfloor} \end{align*}

Similarly the lower limit is sometimes not an integer. In this situation we take the smallest integer greater or equal to the lower limit (ceiling function). \begin{align*} \sum_{k=\frac{m}{2}}^{\frac{n}{2}} a_k=\sum_{k=\color{blue}{\left\lceil\frac{m}{2}\right\rceil}}^{\left\lfloor\frac{n}{2}\right\rfloor} a_k=a_{\left\lceil\frac{m}{2}\right\rceil}+a_{\left\lceil\frac{m}{2}\right\rceil+1}+\cdots+a_{\left\lfloor\frac{n}{2}\right\rfloor} \end{align*}

If the lower limit of the index is equal to $e$, we have consequently \begin{align*} \sum_{k=e}^\infty a_k=\sum_{k=\left\lceil e\right\rceil}^\infty a_k=\sum_{\color{blue}{k=3}}^\infty a_k \end{align*}