# Slowing down the divergence

Is there any unbounded function $f:\mathbb{N} \rightarrow \mathbb{N}$ with $f(n+1) \geq f(n)>1$, such that for every $g:\mathbb{N} \rightarrow \mathbb{N}$ with $g(n+1) \geq g(n)>1$ and $$\sum_{n = 1}^\infty \frac{1}{g(n)} = \infty$$ then $$\displaystyle \sum_{n = 1}^\infty \frac{1}{f(n)g(n)} = \infty$$

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I would guess not. But there may be for every $g:\mathbb{N} \rightarrow \mathbb{N}$ with $g(n+1) \geq g(n)$ and $\sum_{n = 1}^\infty \frac{1}{g(n)} = \infty$ an $f:\mathbb{N} \rightarrow \mathbb{N}$ with $f(n+1) \geq f(n)$, such that $\sum_{n = 1}^\infty \frac{1}{f(n)g(n)} = \infty$. –  Henry Nov 14 '11 at 7:43
Henry, for that one I think you can pick $f$ to be a nested (enough time) logarithm, in which case one would need to know the growth of the partial sums of $1/g(n)$. –  Aleks Vlasev Nov 14 '11 at 7:47
Actually, in Exercise 11, Chapter 3 of Rudin's Principles of Mathematical Analysis, we have that if $a_n > 0$ for every $n\ge 1$, $s_n = \sum_{k=1}^{n}a_k$ then if $\sum a_n$ diverges, so does $\sum a_n/s_n$. So one could pick $a_n = 1/g(n)$ and $f(n) = \sum_{k=1}^{n}1/g(k)$ to see that $\sum 1/(f(n)g(n))$ diverges. –  Aleks Vlasev Nov 14 '11 at 7:57
@Aleks The OP wants an $f$ that has the given property $\forall g$. Your $f$ is $g$-dependent. Am I missing something? –  alex.jordan Nov 14 '11 at 8:09
@alex.jordan You are right, I think Aleks was confirming Henry's claim. –  user1708 Nov 14 '11 at 8:11

The answer is no, that is: for every unbounded nondecreasing positive $(f(n))_n$ there exists a nondecreasing $(g(n))_n$ such that the series $\sum\limits_n1/g(n)$ diverges and the series $\sum\limits_n1/(f(n)g(n))$ converges.

Let us show the equivalent statement that for every nonincreasing positive $(a(n))_n$ such that $a_n\to0$ there exists a nonincreasing $(b(n))_n$ such that the series $\sum\limits_nb(n)$ diverges and the series $\sum\limits_na(n)b(n)$ converges.

To do this, assume without loss of generality that $a(n)\leqslant1$ for every $n$ and, for every $k\geqslant0$, consider $$C_k=\{n\mid 2^{-k-1}<a(n)\leqslant2^{-k}\},\qquad c(k)=|C_k|,\qquad d(k)=\sum\limits_{i\leqslant k}c(i).$$ Let $K$ denote the set of $k$ such that $c(k)\ne0$. For every $k$ in $K$ and every $n$ in $C_k$, let $$b(n)=d(k)^{-1}.$$ Then $\sum\limits_nb(n)=\sum\limits_{k\in K}c(k)/d(k)$ which is infinite(1) hence the series $\sum\limits_nb(n)$ diverges. Since $a(n)\leqslant2^{-k}$ and $c(k)b(n)\leqslant 1$ for every $k$ in $K$ and every $n$ in $C_k$, $\sum\limits_na(n)b(n)\leqslant\sum\limits_{k\in K}2^{-k}$ hence the series $\sum\limits_na(n)b(n)$ converges.

Note (1): For every $k_0$ choose $k_1\geqslant k_0$ such that $d(k_1)\geqslant2d(k_0)$. This can be done since $d(k)\to+\infty$ when $k\to\infty$. Call $K_0$ the set of $k$ such that $k_0\lt k\leqslant k_1$ and $c(k)\ne0$. Then $$\sum\limits_{k\in K_0}c(k)/d(k)\geqslant\sum\limits_{k\in K_0}c(k)/d(k_1)=(d(k_1)-d(k_0))/d(k_1)\geqslant1/2.$$ Thus the rests of the series $\sum\limits_{k\in K}c(k)/d(k)$ do not converge to $0$ hence the series diverges.

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Why is $b$ nonincreasing? –  joriki Nov 14 '11 at 8:47
@joriki, Ouch! Thanks, I modified my post. –  Did Nov 14 '11 at 9:00

No. Let $N_k = \min \{n: f(n) \ge 2^k\}$. I'll leave the case where $N_{k+1} - N_k$ is bounded as an exercise. So assume $N_{k+1} - N_k$ is unbounded. Let $g(n) = \max \{N_{k+1} - N_k: N_k \le n \}$. Note that if $N_{k+1} - N_k > N_{j+1} - N_j$ for all $j < k$, $g(n) = N_{k+1} - N_k$ for $N_k \le n < N_{k+1}$, and $\sum_{N_k \le n < N_{k+1}} \frac{1}{g(n)} = 1$. Since there are infinitely many of these $k$, $\sum_{n=1}^\infty \frac{1}{g(n)} = \infty$. On the other hand, $\sum_{N_k \le n < N_{k+1}} \frac{1}{f(n) g(n)} \le 2^{-k}$, so $\sum_{n=1}^\infty \frac{1}{f(n) g(n)}< \infty$.

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There is no such function $f$. Given $f$, we can construct a counterexample $g$ as follows:

Starting with $g_1=2$ and $n_1=1$, in each step $k$, choose $\Delta_k$ such that $\Delta_k\ge g_k$ and $f(n_k+\Delta_k)\ge2^k$. This is possible since $f$ is unbounded. Now choose $g_{k+1}\ge g_k$ such that $1\le\Delta_k/g_{k+1}\le2$. This is possible, since $\Delta_k/g_k\ge1$. Now choose $n_{k+1}=n_k+\Delta_k$, and $g(n)=g_{k+1}$ for all $n_k\le n\lt n_{k+1}$. This stretch of values adds at least $1$ to the sum of $1/g$ but at most $2^{2-k}$ to the sum of $1/(fg)$, and thus the former diverges while the latter converges.

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