Slowing down divergence 2 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?
 A: 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 $2^{-n}$.
A: 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 $$\sum_n \frac{1}{f(n)+g(n)}=\sum_n \frac{1}{2^n+1} \le \sum_n 2^{-n} <\infty$$
A: 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.
A: 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.
