Let an and bn are strictly increasing to +∞ sequences such that the series ∑1an and ∑1bn are divergent. Is it true that the series ∑1an+bn is also divergent?

At first sight it looks true, so I tried to prove that, but after some failed attempts I start to believe now that there might exist a counterexample. Any thoughts?

**Answer**

For r⩾, let k_r = 2^{2^r}. Let

\begin{align}

a_n &= k_{2r+2} – \frac{1}{n},\text{ for } k_{2r} \leqslant n < k_{2r+2};\\

b_n &= k_{2r+1} – \frac{1}{n},\text{ for } k_{2r-1} \leqslant n < k_{2r+1};

\end{align}

for n \geqslant 4, and choose a_n, b_n fairly arbitrarily for n < 4.

Then

\sum_{n=k_{2r}}^{k_{2r+2}-1} \frac{1}{a_n} > \frac{k_{2r+2}-k_{2r}}{k_{2r+2}} > \frac{1}{2},

so \sum \frac{1}{a_n} diverges. Analogously, \sum \frac{1}{b_n} diverges.

But, we have a_n > b_n for k_{2r} \leqslant n < k_{2r+1}, and b_n > a_n for k_{2r+1} \leqslant n < k_{2r+2}, so

\sum_{n=k_{2r}}^{k_{2r+2}-1} \frac{1}{a_n + b_n} < \frac{k_{2r+1}-k_{2r}}{k_{2r+2}} + \frac{k_{2r+2}-k_{2r+1}}{k_{2r+3}} < \frac{k_{2r+1}}{k_{2r+2}} + \frac{k_{2r+2}}{k_{2r+3}},

and

\frac{k_r}{k_{r+1}} = 2^{2^r-2^{r+1}} = 2^{-2^r} = \frac{1}{k_r},

so

\sum_{r=1}^\infty \frac{1}{k_r} < \infty

and

\sum_{n=1}^\infty \frac{1}{a_n+b_n}

converges.

**Attribution***Source : Link , Question Author : CuriousGuest , Answer Author : Daniel Fischer*