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Let $f(x)$ and $g(x)$ be positive nondecreasing functions such that $ \sum_{n>1} \frac1{f(n)} \text{ and } \sum_{n>1} \frac1{g(n)} $

diverges.

(Why) must the series $$\sum_{n>1} \frac1{g(n)+f(n)}$$ diverge?

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    $\begingroup$ Because the name of 1/(f+g) always has at least as many Ws in it as the names of 1/f and 1/g combined. $\endgroup$
    – joriki
    Commented Nov 15, 2011 at 9:39

4 Answers 4

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No, $\sum\frac{1}{f(n)+g(n)}$ need not diverge. We define such a pair $f,g$ inductively. Let $f(1) = g(1) = 2$.

Suppose $f(n)$ and $g(n)$ have been defined for $1\leq n\leq N$. Let $L$ be the largest value taken by $f(n)$ or $g(n)$ for $1\leq n\leq N$. Define $f(N+1) = f(N+2) = \cdots = f(N + L) = L$ and $g(N+i) = 2^iL$ for $1\leq i\leq L$. Define $g(N+L+1) = \cdots = g(N+L+2^LL) = 2^LL$ and $f(N+L+i) = 2^{L+i}L$ for $i\leq 1\leq 2^LL$. The extends the domain of definition of $f(n)$ and $g(n)$ to $1\leq n\leq N+L+2^LL$. Repeat to define $f$ and $g$ on all of $\mathbb{N}$.

Note that $f$ and $g$ are positive and nondecreasing. In each stage of this process there are $L$ times when $f$ takes the value $L$ and $2^LL$ times when $g$ takes the value $2^LL$. Therefore $\sum_{n=N+1}^{N+L+2^LL} \frac{1}{f(n)} > 1$ and $\sum_{n=N+1}^{N+L+2^LL} \frac{1}{g(n)} > 1$: at least $1$ is added to the sum of the reciprocals of each series at each step. Since we repeat this process ad infinitum, adding at least $1$ to each series at each step, $\sum_{n=1}^\infty \frac{1}{f(n)}$ and $\sum_{n=1}^\infty \frac{1}{g(n)}$ both diverge.

Note that by construction, $\max(f(n+1),g(n+1)) = 2\max(f(n),g(n))$ for all $n$. Since $f(1)=g(1)=2$, we have $\max(f(n),g(n))=2^n$ for all $n$. Therefore $\frac{1}{f(n)+g(n)}\leq \frac{1}{\max(f(n),g(n))} = 2^{-n}$ for all $n$, so $\sum_{n=1}^\infty\frac{1}{f(n)+g(n)}$ converges by comparison with the geometric series $2^{-n}$.

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    $\begingroup$ This looks good to me, but someone gave it a down vote. Doesn't it work? Basically you take a sequence $A$ that grows fast enough for $\sum(1/A)$ to converge, and write it as $A=B+C$ where $B$ and $C$ are flat for long enough stretches for $\sum(1/B)$ and $\sum(1/C)$ to diverge. $\endgroup$ Commented Nov 16, 2011 at 6:39
  • $\begingroup$ So what would happend if f and g were strictly increasing? $\endgroup$ Commented Nov 16, 2011 at 8:14
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    $\begingroup$ @Holowitz: If $f,g$ build a counterexample being non-decreasing then you can think about taking $f^*_n = f_n+\varepsilon_n$ and $g^*_n = g_n+\delta_n$ to be strictly increasing sequences. For this case you take $\varepsilon,\delta$ to be positive sequences convergent to zero. I believe that $\varepsilon_n = \delta_n = 2^{-n}$ should work $\endgroup$
    – SBF
    Commented Nov 16, 2011 at 8:27
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    $\begingroup$ @Gortaur: that won't work. For the $\epsilon_n$ and $\delta_n$ you chose, where $f_n$ were constant in $n$ your $f^*_n$ is actually strictly decreasing. For Noah's construction, however, you can choose the perturbation $\epsilon_n = \delta_n = \log \log \log n$ which is slowly increasing such that your $f^*_n$ and $g^*_n$ are within a factor of 2 times the original $f_n$ and $g_n$, so still diverges. $\endgroup$ Commented Nov 16, 2011 at 9:18
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    $\begingroup$ So, you essentially have $$ {f(n) : 2^{\hphantom {2}}\ 2^{\hphantom{2}} \ \ \overbrace{2^4\ 2^5\ \cdots\ 2^{11}}^{8 \rm\ terms} \ \ 2^{12}\ 2^{12}\cdots \ 2^{12}\ 2^{12 \hphantom{+11}} \ \cdots } \atop {g(n) : 2^2\ 2^3 \ \ 2^3\ 2^3\ \cdots\ 2^{3\hphantom{1} }\ \ \ \underbrace{2^{12}\ 2^{13} \ 2^{14} \cdots 2^{2^{12} +11}}_{2^{12}\rm terms }\cdots} $$ Very nice! $\endgroup$ Commented Nov 16, 2011 at 9:53
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Maybe obvious remark, but too long to put as a comment (honestly, I've tried). I put it here until the moment the problem will be solved.

I was trying to go the same way as Gerry suggested (I hope he meant $\min$ in his answer), i.e. $a_n=\frac{1}{f_n}$, $b_n = \frac{1}{g_n}$ are non-increasing positive sequences such that $$ \sum_na_n = \sum_n b_n = \infty $$ and the question is if $$ \sum_n c_n = \sum_n\min\{a_n,b_n\}=\infty. $$ It's only sufficient of course, so if one will find a counterexample for this problem it's not necessary a solution for the original problem.

However, for a counterexample to the original problem (if one exists) we should have $a_n\geq b_n$ and $b_n\geq a_n$ infinitely many times, or equivalently $f_n\geq g_n$ and $g_n\geq f_n$ infinitely many times. If it does not hold than the residual of sum $\sum_n c_n$ consists either only of $a_n$ or of $b_n$ and hence diverges.

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  • $\begingroup$ Abel's Theorem is (to me): If $\{a_n\}$ is positive and non-increasing and $\sum a_n$ converges, then $na_n\rightarrow 0$. This actually goes by another name, but I don't recall what it is. I'll look it up... $\endgroup$ Commented Nov 15, 2011 at 17:14
  • $\begingroup$ @David: I believe in your previous comment you meant that $n\cdot c_n\to 0$ implies that at least one of $\sum a_n$ and $\sum b_n$ converge? $\endgroup$
    – SBF
    Commented Nov 15, 2011 at 17:15
  • $\begingroup$ Yes. I thought I could show this, but not sure now. Sorry for the confusion. The theorem I referred to earlier (stated above) is Pringsheim's Theorem according to some, but was discovered originally by Abel. $\endgroup$ Commented Nov 15, 2011 at 17:22
  • $\begingroup$ @Gortaur: I've added a follow-up as a separate 'answer'. $\endgroup$ Commented Nov 18, 2011 at 8:55
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The hypothesis that the sequences are nondecreasing is essential since otherwise there is the following counterexample (which I believed to be an answer because I missed the hypothesis):

Let $f(n)=2^n$ for $n$ odd and $f(n)=1$ for $n$ even, and let $g(n)=2^n$ for $n$ even, $g(n)=1$ for $n$ odd. Then both $\sum \frac1{f(n)}$ and $\sum \frac1{g(n)}$ diverge because they contain 1 infinitely often, but $$\sum_n \frac{1}{f(n)+g(n)}=\sum_n \frac{1}{2^n+1} \le \sum_n 2^{-n} <\infty$$

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    $\begingroup$ The op wants nondecreasing functions. $\endgroup$ Commented Nov 15, 2011 at 9:44
  • $\begingroup$ OK, I missed that. $\endgroup$
    – Florian
    Commented Nov 15, 2011 at 10:26
  • $\begingroup$ @Florian: you've also missed 2 downvotes ) $\endgroup$
    – SBF
    Commented Nov 15, 2011 at 21:53
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    $\begingroup$ The downvote is not from me, but why are people still upvoting this? It does not answer the question at all... $\endgroup$
    – Listing
    Commented Nov 15, 2011 at 22:40
  • $\begingroup$ Maybe some people give upvotes to reward effort. $\endgroup$ Commented Nov 16, 2011 at 6:09
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This is a follow-up to Gortaur's "answer" (and isn't a real answer either).

If $a_n$ and $c_n$ are as above, then to show that $\sum_nc_n=\infty$, I believe that it is sufficient to show that if $\;1=i_0< i_1<i_2<\dots\;$ is any infinite sequence of indexes, then $\sum_jd_j=\infty$ where $$ d_j\;=\;a_{i_k} \;\;\textrm{where}\;\; i_{k-1}\leqslant j<i_k. $$ This reduces the question to one about a single non-increasing positive sequence. I haven't been able to find a counter-example to this.

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