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I had asked this question sometime ago here. Now I have a question which I think is related to it.

Let $f$ be an increasing function (continuous, of course!) with $f(1)=0$.

Consider the sequence $s_{n}= ( \sum\limits_{k=1}^{n} f(k) - \int\limits_{1}^{n} f(x) dx )$.

When does $s_{n}$ converge?

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This question seems far too general to me, but is probably relevant. – Qiaochu Yuan Aug 27 '10 at 0:47
@Qiaochu Yuan: Thanks for the link +1 :). – anonymous Aug 27 '10 at 1:26
up vote 6 down vote accepted

Qiaochu was on the right track to use an integral-to-sum formula, but it sounds like you want the Abel-Plana summation formula:

$$\lim_{n\to\infty}\left(\sum_{k=m}^n f(k)-\int_m^n f(u)\mathrm{d}u\right)=\frac{f(m)}{2}-\int_{-\infty}^\infty \left(\frac{|t|}{\exp(|2\pi t|)-1}\right)\left(\frac{f(m+it)-f(m-it)}{2it}\right)\mathrm{d}t$$

This is used for instance to evaluate the Stieltjes constants. If the expression on the right hand side is convergent, then it is equivalent to the left hand side.

Adendum for Chandru:

Definitely $f(z)$ should be analytic, or at least analytic in the region where $\Re z\geq m$. Per Henrici's "Applied and Computational Complex Analysis", the additional conditions are

$$\lim_{t\to\infty}f(u\pm it)=0$$

uniformly in $u$, and that

$$\lim_{t\to\infty}|f(u\pm it)|\exp(\mp 2\pi t)=0$$

uniformly in $u$.


For those scratching their head on just how Abel-Plana and Euler-Maclaurin are connected, the identity

$$\int_{-\infty}^\infty \left(\frac{|t|}{\exp(|2\pi t|)-1}\right)|t|^{2n-2}\mathrm{d}t=\frac{|B_{2n}|}{2n}$$

might be of interest.

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Mangaldan: So when is the expression on the right hand side convergent! – anonymous Aug 27 '10 at 1:59
@J. Mangaldan: I don't see the need for analyticity. For example, let f be piecewise linear with vertices at (n, 1-exp(-(n-1))) for n = 1, 2, ... . It increases continuously with f(1)=0. The series is obviously bounded and increasing, so the limit exists (and equals 1/2), but f is not analytic. – whuber Aug 27 '10 at 3:04
Hmm... the proof of Abel-Plana in Henrici started with the analyticity assumption. Unfortunately I've not sufficient complex analysis machinery to show that a weaker condition than analyticity can be used. – J. M. Aug 27 '10 at 3:14
@J. M. This is a beautiful formula you have shared with us; thank you. My point above is that it (of course) requires that f be analytic, but the original problem statement imposes no such restriction. Thus this formula can identify some convergent situations but not necessary all. – whuber Dec 14 '10 at 20:42

The question implicitly asks for a "simpler" or "more interesting" criterion for convergence. I doubt there is one. Intuitively, f can do almost anything between (n, f(n)) and (n+1, f(n+1)) provided it is increasing. Thus the terms of (s(n)) can be anything. Convergence therefore is determined by fairly arbitrary properties of f(n) as n becomes arbitrarily large. If you don't severely restrict f--e.g., require it to be analytic or bounded and concave or something like that--you shouldn't expect to find any simpler answer than "(s(n)) converges when it converges."

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The following theorem (which I read in Number theory: algebraic numbers and functions By Helmut Koch)

$$ \sum\limits_{n=1}^{N} g(n) = g(1) + \int\limits_{1}^{N} g(x) \mathrm{d}x + \int\limits_{1}^{N} (x - [x]) g'(x) \mathrm{d}x $$

tells us that the difference converges when $$\int\limits_{1}^{N} (x - [x]) g'(x) \mathrm{d}x$$ does.

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Note that $$s_{n+1}-s_n =f(n+1)- \int_{n}^{n+1} f(x) dx \,.$$

Let $$a_n:= f(n+1)- \int_{n}^{n+1} f(x) dx = \int_n^{n+1} [f(n+1)-f(x)] dx \,.$$

Then your sequence is exactly the sequence of partial sums of the positive series $$ \sum_n a_n \,.$$

When is this convergent? It is convergent if and only if $a_n \to 0$ "fast enough".

Basically your question asks: under what conditions does $\int_n^{n+1} [f(n+1)-f(x)] dx$ converge to zero fast enough so that the corresponding series is absolutely convergent?

Here is a simple condition, but it is probably completelly useless for practical applications: Let $c_n \in (n,n+1)$ be so that $\int_n^{n+1}f(x)dx =f(c_n)$. Then your sequence is convergent if and only if $\sum _n [ f(n+1)-f(c_n)]$ is convergent...

P.S. Probably a better question to ask is the following:

Define $g_n : [0,1] \rightarrow R$ by $g_n(x) = f(n+1)-f(n+1-x)$. Then $g_n$ is continuous on $[0,1]$, increasing and $g_n(0)=0$. Keep in mind that any such $g_n$ lead to an unique $f$ which verifies your conditions.

Then, the question you ask becomes equivalent to the following:

Let $g_n : [0,1] \rightarrow R$ be continuous, increasing and $g_n(0)=0$. Under what extra conditions is

$$\sum_n \left( \int_0^1 g_n(x) dx \right) $$ convergent?

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