Are alternating series just derivative series?

Question

I have a thought I cannot get out of my head but also cannot seem to prove or disprove satisfyingly. Please help me clarify my thoughts.

Consider an alternating series $\sum{({-1})^{n+1}u_n}$ where $u_n$ is positive. We could group two by two this series and have: $$\sum_{n\ge0}{({-1})^{n+1}u_n}=\sum_{n\ge0}{\left(u_{2n+1}-u_{2n}\right)}.$$ I’ll now change notations to explicit calculus: let $f(n)=u_n$, now: $$\sum_{n\ge0}{({-1})^{n+1}u_n}=\sum_{n\ge0}\left(f(2n+1)-f(2n)\right).$$ Now, as $n$ goes to infinity, my gut says I’m almost looking at the derivative of $f(x)$: $$\frac{f(x+1)-f(x)}{1},$$ where $1$ is a really tiny $h$ when compared to $x$ going to infinity. So I’m really tempted to conclude that: $$\sum_{n\ge0}{({-1})^{n+1}u_n}=O\left(\sum_{n\ge0}{f'(2n)}\right).$$ I tried selecting a specific function like for example $f(x)=\frac{1}{x}$ and of course: $$f(x+1)-f(x)=-\frac{1}{x^2}+o\left(\frac{1}{x^2}\right)=f'(x)+o(f'(x)).$$ Beyond that though, I can’t really make a rigorous proof of my intuition being correct, or that it is fundamentally wrong and should be rejected. One last thing I need to mention, obviously I grouped 2 by 2 my alternating series, so I should heed at some point the grouping theorem conditions: my grouping scheme should be bounded independently of $n$ (in my case, it’s 2 by 2, so I meet this requirement) and $u_n$ should converge to 0.

Does anyone have any idea to help me find closure? I’m not sure that if this result is correct by the way, that it would be useful to study an alternating series: I wonder if it would be simpler to compute the derivative of $f$ and look at it to see if the series converges.

Answer 1

Suppose $f \colon [0, \infty) \to [0, \infty)$ is continuously differentiable, weakly decreasing, and $\lim_{x \to \infty} f(x) = 0$. Let $u_n = f(n)$. It seems to me you wish to compare $$\sum_{n=0}^{2N+1} (-1)^{n+1} u_n \qquad\text{vs.}\qquad\sum_{n=0}^N f'(2n).$$ By the mean value theorem, for each $n$ we have some $\xi_n \in [2n, 2n+1]$ such that $u_{2n+1} - u_{2n} = f'(\xi_n)$. Thus $$\sum_{n=0}^{2N+1} (-1)^{n+1} u_n = \sum_{n=0}^N f'(\xi_n)$$ and the question becomes how the errors between $f'(\xi_n)$ and $f'(2n)$ accumulate.

Let us now additionally assume $|f'|$ is monotonically decreasing, necessarily with $\lim_{x \to \infty} f'(x) \to 0$. Then $f'(2n) \leq f'(\xi_n) \leq f'(2n+1) \leq f'(2(n+1))$, so $$\sum_{n=0}^N f'(2n) \leq \sum_{n=0}^N f'(\xi_n) \leq \sum_{n=1}^{N+1} f'(2n).$$ Thus $$\boxed{0 \leq \sum_{n=0}^{2N+1} (-1)^{n+1} u_n - \sum_{n=0}^N f'(2n) \leq f'(2N+2) - f'(0).}$$ In particular the difference is $O(1)$.

Surely one could cook up $f$ where $|f'|$ is not monotonically decreasing but instead oscillates enough for the errors to accumulate appreciably. The above is merely a prototype of an argument which chooses a few reasonable conditions to conclude something like what you seem to be after.