# If $\sum a_n$ and $\sum b_n$ converge then is it true that$\sum\max\{a_n,b_n\}$ converges?

Let $$\sum a_n,\sum b_n$$ be convergent series and let $$A_n,B_n$$ be the $$n\text{th}$$ partial sums of $$(a_n),(b_n)$$ respectively. Let $$c_n=\max\{a_n,b_n\}$$ and let $$C_n$$ be the $$n\text{th}$$ partial sum of the sequence $$(c_n)$$. If it were true that $$C_n=\max\{A_n,B_n\}$$ then we would conclude that $$\sum\max\{a_n,b_n\}=\max\left\{\sum a_n,\sum b_n\right\}$$ due to this, but obviously it's not.

We know that for every $$x,y\in\mathbb{R}$$ we have $$\max\{x,y\}=\frac{x+y+|x-y|}{2}.$$ From this we can see that if $$\sum a_n,\sum b_n$$ are absolutely convergent then so is $$\sum\max\{a_n,b_n\}=\sum\frac{a_n+b_n+|a_n-b_n|}{2}$$ because $$0\leq|a_n-b_n|\leq|a_n|+|b_n|$$ and $$\sum |a_n|+|b_n|$$ is convergent.

If $$\sum a_n,\sum b_n$$ converge (not necessarily absolutely) then can we say that $$\sum\max\{a_n,b_n\}$$ converges?

• what exactly is $\sum\max\{a_n+b_n\}$?
– SBF
Aug 21, 2023 at 9:19
• @SBF My bad. Fixed. Aug 21, 2023 at 9:21
• Just consider a case where $0 < a_n < b_n$ for $n>1$ and $0 < b_1 < a_1$ Aug 21, 2023 at 9:21
• This theorem is not even true for finite sums, take $a_1=1, b_1 =3, a_2=2, b_2=1$ for example.
– Eric
Aug 21, 2023 at 9:22
• See the counterexample in the answer to this post. Aug 21, 2023 at 9:36

No, the conclusion cannot be made. Simply take a convergent but not absolutely convergent series with alternating signs as $$(a_n)$$ and $$b_n=-a_n$$.
For example, use for $$n = 1,2,\ldots$$
$$a_n = (-1)^n\frac1n,\; b_n=-a_n = (-1)^{n+1}\frac1n.$$
Then $$\sum_{n=1}^\infty b_n = \ln 2, \sum_{n=1}^\infty a_n = -\ln 2$$, but we get $$\max\{a_n,b_n\}=\frac1n$$, so $$\sum_{n=1}^\infty \max\{a_n,b_n\}$$ is the divergent harmonic series.
If you don't know that $$\sum_{n=1}^\infty b_n = \ln 2$$, it is at least easy to see that the sum converges, due to the alternating series test.