# A criterion for series convergence?

I have a conjecture regarding series convergence that feels like it would be a useful tool to me if I could prove it, but I have been unable to prove or disprove it.

Let $\sum a_n$ be a series of nonnegative terms. Define $\lambda(N)$ to be the number of terms of the series that are greater than $1/N$.

Conjecture: If $\lambda(N)=O(N^\alpha)$, with $0<\alpha<1$, then $\sum a_n$ converges.

The conjecture is based on the fact that this is true for series of the form $\sum 1/n^k$, $k$ constant (let $\alpha = 1/k$), and my intuition that the hypothesis of the conjecture is enough to make $\sum a_n$ "sufficiently similar to or bounded by" such a series.

Can you offer a counterexample or point me toward an idea for a proof?

(I tried to bound $\Delta\lambda(N) = \lambda(N+1)-\lambda(N)$ from the assumption $\lambda(N)=O(N^\alpha)$, since possibly excluding a finite number of terms, the series sum is bounded above by $\sum \Delta\lambda(N)/N$; but I couldn't see how to do this.)

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Let $M>0$ be such that $\lambda(N)<M\cdot N^\alpha$ for all $N$. Summation by parts yields

\begin{align*} \sum_{N=1}^{n-1}\frac{1}{N}(\lambda(N+1)-\lambda(N)) &=\frac{\lambda(n)}{n}-\lambda(1)+\sum_{N=1}^{n-1}\lambda(N+1)\left(\frac{1}{N}-\frac{1}{N+1}\right)\\ &\leq\frac{\lambda(n)}{n}+\sum_{N=1}^{n-1}\frac{\lambda(2N)}{N^2}\\ &\leq\frac{M}{n^{1-\alpha}}+M\cdot2^\alpha\sum_{N=1}^{n-1}\frac{1}{N^{2-\alpha}} \end{align*}

for all $n$, so $$\sum_{N=1}^\infty\frac{1}{N}(\lambda(N+1)-\lambda(N))\leq M\cdot2^\alpha\sum_{N=1}^\infty\frac{1}{N^{2-\alpha}}<\infty.$$

And as you mentioned, $\sum a_n\leq c+\sum \Delta\lambda(N)/N$ for some constant $c$ corresponding to the finite number of terms larger than $1$.

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I suppose the unstated point is that $\sum a_n$ can be bounded by that first sum? – Gerry Myerson Aug 4 '11 at 6:14
Gerry, thanks, now I've stated it. – Jonas Meyer Aug 4 '11 at 6:17

This is summation by blocks in disguise.

For every $k\ge0$, the sum of the terms $a_n$ such that $2^{-(k+1)}\le a_n\le 2^{-k}$ is at most the number of terms times a common upper bounds of these, hence at most $\lambda(2^{k+1})$ times $2^{-k}$. It happens that the number $\lambda(1)$ of terms $a_n$ such that $a_n\ge1$ is finite (1), hence it suffices to consider the sum of the terms $a_n$ such that $a_n\le1$. This sum is at most $$\sum_{k=0}^{+\infty}\sum_na_n\cdot[2^{-(k+1)}\le a_n\le 2^{-k}]\le\sum_{k=0}^{+\infty}\lambda(2^{k+1})2^{-k}.$$ Now, if $\lambda(N)\le c\cdot N^a$ for every $N$ large enough, then $\lambda(2^{k+1})\le c\cdot2^{a(k+1)}$ for every $k$ large enough and the last series above is controlled by the series $$\displaystyle2^ac\cdot\sum_k2^{-(1-a)k},$$ which converges since the hypothesis that $a<1$ implies that $2^{-(1-a)}<1$.

(1) The proof is simple: $\lambda(1)\le\lambda(N)$ for every $N\ge1$ and the hypothesis that $\lambda(N)=O(N^a)$ implies that $\lambda(N)$ is finite for $N$ large enough.

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+1 wish I could accept both answers! I was thinking that the assumption that $\lambda(N) = O(N^\alpha)$ rules out the possibility of $\lambda(N)$ ever being infinite. I want this because I want the conjecture to include this case: if $\lambda(N)$ is ever infinite, then it is certainly not $O(N^\alpha)$ for $\alpha \leq 1$, and sure enough the series diverges. – Ben Blum-Smith Aug 4 '11 at 12:55
This is both right and wrong... In general, the hypothesis that two sequences $(a_N)$ and $(b_N)$ are such that $a_N=O(b_N)$ concerns only their asymptotics, in the sense that, for every fixed finite $n$, if one modifies some or every $a_N$ and $b_N$ for $N\le n$ one does not change the conclusion. But in your case, the sequence $(\lambda(N))$ is nondecreasing hence if $\lambda(n)$ was infinite for a given $n$, then $\lambda(N)$ would be infinite for every $N\ge n$ as well and the hypothesis $\lambda(N)=O(N^a)$ would not hold. (I missed that fact in my answer, so I will modify it.) – Did Aug 4 '11 at 13:22

Another way of thinking about it:

Let $f(n):=a_n$ and extend $\lambda(N)$ to $\mathbb R$ by defining $\lambda(t)=\#\{n:f(n)>1/t\}$ (here $\#$ means cardinality). Define $F_f(t)=\lambda(1/t)=\#\{n:f(n)>t\}$; if we think of the set of positive integers as a measure space, with the counting measure, then the function $F_f(t)$ is the distribution function of $f$, and $$\sum_{n=1}^{\infty}f(n)=\int_0^{\infty}F_f(t)\,\mathrm dt.$$ Your hypothesis is that $F_f(1/N)=O(N^{\alpha})$. In particular, there are only finitely many values of $f$ that are $>1$, and $f$ is bounded.

Since $\lceil t^{-1}\rceil^{-1}\leq t<2\lceil t^{-1}\rceil^{-1}$ for $t\leq 1$, we can convert the bound on $F_f(1/N)$ to a bound on $F_f(t)$: $$F_f(t)\leq F_f(\lceil t^{-1}\rceil^{-1})\leq C\lceil t^{-1}\rceil^{\alpha}\leq C2^{\alpha} t^{-\alpha}\qquad(0<t\leq1),$$ for some constant $C$. That is to say, $F_f(t)=O(t^{-\alpha})$. (This is the same as saying that $f$ is in weak $\ell^{\alpha}$.) Now we can just approximate the integral above: If $M$ is a bound for $f$, then $$\int_0^{\infty}F_f(t)\,\mathrm dt = \int_0^MF_f(t)\,\mathrm dt \leq C'\int_0^1t^{-\alpha}\,\mathrm dt + C'MF_f(1) <\infty,$$ since $0<\alpha<1$.

In fact, your observation can be generalized: Let $L^{p,\infty}(\mu)$ denote weak $L^p(\mu)$, the space of all functions $f:(X,\mu)\to\mathbb C$ satisfying $$\|f\|_{p,w}:=\sup_{t>0}\,\,{t\,\mu(\{x:|f(x)|>t\})^{1/p}}<\infty.$$ If $f\in L^{p,\infty}(\mu)\cap L^{q,\infty}(\mu)$, where $0<p<q\leq\infty$, then $f\in L^r(\mu)$ for all $r\in (p,q)$. (See Grafakos, Classical Fourier Analysis, pp. 8-9 for a more precise result; also, Terry Tao's blog has some interesting information in connection with this topic in general.)

To apply this generalization to your specific case, you can take $p=\alpha$, $q=\infty$ (because $\{a_n\}$ is bounded), and $r=1$ to conclude that $\{a_n\}\in \ell^1$.

(I wanted to say something about this yesterday, but I couldn't think of the appropriate generalization. Being new to the subject, I also wanted to check with a professor of mine to ensure that the above analogy was apt. Please let me know if there are errors.)

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+1 I'm too new to measure theory to be able to check your details but thank you for taking the time to approach the question from this point of view. – Ben Blum-Smith Oct 3 '11 at 23:23