# $\sum_{n=1}^{\infty}\frac{1}{a_n}=\infty$ $\sum_{n=1}^{\infty}\frac{1}{b_n}=\infty$ but $\sum_{n=1}^{\infty}\frac{1}{a_n+b_n}=c, c\in R$ [closed]

The following question is inspired from: Defining the slowest divergent series.

Let $a_n$ and $b_n$ be two strictly increasing sequences of natural numbers,with $\sum_{n=1}^{\infty}\frac{1}{a_n}=\infty$ and $\sum_{n=1}^{\infty}\frac{1}{b_n}=\infty$

When I was an undergraduate student, i had in mind the ''conjecture'' that using the above sequences it is impossible to have $\sum_{n=1}^{\infty}\frac{1}{a_n+b_n}=c, c\in R$.

Surprisingly enough, i believe we can find a counterexample: We can find two sequences $a_n,b_n$ with $\sum_{n=1}^{\infty}\frac{1}{a_n}=\infty,$ $\sum_{n=1}^{\infty}\frac{1}{b_n}=\infty$ but $\sum_{n=1}^{\infty}\frac{1}{a_n+b_n}=1$ (or some other constant, but this is a specific example)

Let $a_n+b_n=2^n$

The idea is to increase the sum of the first series by a number aproximately $ln2$ between the

$2^{2^{.^{.{^2}}}}$ ($n-2$ exponents) and $2^{2^{.^{.{^2}}}}$ ($n$ exponents) term. The same idea will also run for the second series.

$a_1=b_1=1\\a_2=b_2=2\\a_3=5,b_3=3\\a_4=12,b_4=4$

because $a_4$ is ''too big'' we will increase $a_n$ only by $1$ for the next $12$ terms, and increase significally $b_n$ in the following way:

$a_5=13 , b_5=19,a_6=14, b_6=50 ...$ and so on, until we reach $a_{16}=24, b_{16}=65512$$.$ Again we will switch the growth of $b_n$ and $a_n,$ adding only $1$ to $b_n$ for the next $65512$ terms but increase significally $a_n$ in order to sum $2^n$. So, we will reach at $2^{2^{2^{2}}}=2^{65536}=a_{65536}+b_{65536},$ with $a_{65536}=2^{65536}-131024$ and $b_{65536}=131024$ and then continue in the same way.

Both $a_n, b_n$ are strictly increasing and it is easy to see that when we reach $2^{2^{.^{.{^2}}}}$ ($2n$ exponents) both $\sum{\frac{1}{a_n}}$ and $\sum{\frac{1}{b_n}}$ will be at least $nln2$ and so, conclude that both series reach infinity. Also, $\sum \frac{1}{a_n+b_n}=\sum\frac{1}{2^n}=1$

Note:we use $''ln2''$ because $\frac{1}{n+1}+...+\frac{1}{2n}$ has limit $ln2$

(I tried really much to present this question in a more simple way without using so many numericall details,but i could not see any better way. Please feel free to give a simpler version of the above if you wish)

Question:Is this a known fact? I checked Wolfram, Wikipedia, MO but i didn't find anything. It is also mentioned in the above question that ''This question was considered by du Bois-Reymond in 1870 and he came to some conclusions about convergence classes being linearly ordered that were not well substantiated'' but i couldn't find this ''consideration'' of du Bois-Reymond.

Can anybody give me any references?

Motivation: Let's go again to the question at the beggining:We could define a set $A$ of pairs of $(\sum{\frac{1}{a_n}}$,$\sum{\frac{1}{b_n}})$ both being divergent, but with the property that $\sum_{n=1}^{\infty}\frac{1}{a_n+b_n}$ is convergent. I believe that this will define the ''pair of slowest divergent series''

(this is already posted there)

## closed as off-topic by Michael Renardy, Andreas Thom, David White, Andrés E. Caicedo, Kevin VentulloOct 6 '13 at 22:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Michael Renardy, Andreas Thom, David White, Andrés E. Caicedo, Kevin Ventullo
If this question can be reworded to fit the rules in the help center, please edit the question.

• See math.stackexchange.com/questions/82309/… for a similar construction as yours. – user40963 Oct 6 '13 at 20:48
• @yup i had no idea about this.this seems to answer the case where f(n) and g(n) are non decreasing but not strictly increasing as mentioned above.but is a similar construction i suppose. – Konstantinos Gaitanas Oct 6 '13 at 20:52