How to apply the Abel-summation on this series: $\sum_{k=1}^{\infty}((-1)^{k-1} \cdot k) = 1 - 2 + 3 - 4 + ...$

Question

For a project I'm planning on using the Abel-summation to find a solution for the series I'm currently working with. These series being $$\sum_{k=1}^{\infty}((-1)^{k-1} \cdot k) = 1 - 2 + 3 - 4 + ...$$ I know the answer is supposed to be $\frac{1}{4}$, and I figured out how to transform $$\zeta{(s)} = \sum_{k=1}^\infty \frac{1}{k^s}$$ into $$\zeta{(s)} = \frac{\sum_{k=1}^{\infty}(\frac{(-1)^{k-1}}{k^s})}{1 - 2^{1-s}}$$ thus writing $\zeta{(-1)}$ as $$\zeta{(-1)} = \frac{\sum_{k=1}^{\infty}(\frac{(-1)^{k-1}}{k^{-1}})}{1 - 2^{2}} = \frac{\sum_{k=1}^{\infty}((-1)^{k-1} \cdot k)}{1 - 4} = \frac{\sum_{k=1}^{\infty}((-1)^{k-1} \cdot k)}{-3}$$ the only thing I'm having trouble with is calculating $$\sum_{k=1}^{\infty}((-1)^{k-1} \cdot k) = 1 - 2 + 3 - 4 + ...$$ Wikipedia told me that I should use the Abel-summation method, since it's known as a "strong" method, so I figured $a_n = 1$ and $\phi{(x)} = x(-1)^{x-1}$ thus $A(x) = floor(x)$ (don't know how to implement that over here in math stackexchange) resulting into $$\sum_{k=1}^{\infty}((-1)^{k-1} \cdot k) = \lim_{n \to \infty} floor(n) \cdot (-1)^{n-1} \cdot n - \int_{1}^{n} floor(u)\cdot -i(-1)^{u} \cdot (\pi u - i) du$$ the only problem is I don't know how to continue this. I'm not even sure I'm applying the Abel-summation in a right way. Could someone please help me out?

Answer 1

There are two quite different methods that are often called "Abel summation". Both are powerful and important in their respective realms. One is the method of summation by parts - often in the specific form of Abel's sum formula:

If $(a_n)_{n \geqslant 1}$ is a sequence of complex numbers, and $f \colon [1,+\infty) \to \mathbb{C}$ a continuously differentiable function, then we have $$\sum_{n \leqslant x} a_n f(n) = A(x)f(x) - \int_1^x A(t)f'(t)\,dt$$ where $A(x) = \sum_{n \leqslant x} a_n$.

This can of course be generalised, we don't need to start the sum at $1$, and $f$ need not necessarily be continuously differentiable. Anachronistically, Abel's sum formula is a case of the integration by parts formula for Stieltjes integrals. It is widely used in analytic number theory, but it is not very useful to sum divergent series. That is the domain of the other method called "Abel summation":

Let $(a_n)_{n \geqslant 0}$ a sequence of complex numbers such that the power series $$\sum_{n = 0}^{\infty} a_nz^n$$ has radius of convergence at least $1$. Then $\sum_{n = 0}^{\infty} a_n$ is called Abel summable if $$\lim_{x \to 1^-} \sum_{n = 0}^{\infty} a_n x^n$$ exists. This limit is then the Abel-sum of the series.

By Abel's theorem, a convergent series is Abel summable and its Abel-sum is the limit of the sequence of partial sums. But there are many divergent series that are Abel summable, for example the series you ask about. This method can be extended and generalised, as mentioned in the wikipedia article on divergent series.

To find the Abel-sum of the divergent series $$\sum_{k = 1}^{\infty} (-1)^{k-1}\cdot k$$ it is convenient to use an index shift and write $k = n+1$. Then we must look at the function defined on the open unit disk by $$f(z) = \sum_{n = 0}^{\infty} (-1)^n(n+1)z^n\,.$$ It is easy to see that on the open unit disk we have $$f(z) = -\frac{d}{dz}\sum_{n = 0}^{\infty} (-z)^n = -\frac{d}{dz}\frac{1}{1+z} = \frac{1}{(1+z)^2}\,.$$ Thus $f$ is the restriction of a rational function whose only pole is at $-1$, and we see $$\text{A-}\sum_{k = 1}^{\infty} (-1)^{k-1}\cdot k = \lim_{x \to 1^-} \frac{1}{(1+x)^2} = \frac{1}{(1+1)^2} = \frac{1}{4}\,.$$

It may be worthwhile to explicitly note that Abel summation cannot be used to sum the divergent series $$\sum_{n = 1}^{\infty}1 \qquad\text{and}\qquad \sum_{n = 1}^{\infty} n$$ since the functions given by the corresponding power series have a pole at $1$. These (as well as the series in the question) can be summed for example by analytic continuation of Dirichlet series.