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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?

Hilbert
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  • what exactly is $\sum\max{a_n+b_n}$? – SBF Aug 21 '23 at 09:19
  • @SBF My bad. Fixed. – Hilbert Aug 21 '23 at 09:21
  • Just consider a case where $0 < a_n < b_n$ for $n>1$ and $0 < b_1 < a_1$ – Henry Aug 21 '23 at 09:21
  • See the counterexample in the answer to this post. – terran Aug 21 '23 at 09:36
  • $\sum\max{a_n,b_n}=\max\left{\sum a_n,\sum b_n\right}$ is clearly false: on the LHS, you take the largest number between $a_n$ and $b_n$ for every $n$ and then sum all those largest numbers, while on the RHS, you sum the two series and keep the largest result. So $\sum\max{a_n,b_n}\ge\max\left{\sum a_n,\sum b_n\right}$ (always true for finite sums, and assuming convergence of the three series involved for series) and the inequality can be strict. – Taladris Aug 21 '23 at 10:15
  • "$\sum |a_n|+|b_n|$ is convergent." This should be proved. And "$\sum |a_n|+|b_n|$ is convergent. If $\sum a_n,\sum b_n$ converge (not necessarily absolutely)" is actually false: it is not hard to show (using Comparison Test) that $\sum (|a_n|+|b_n|)$ is convergent if and only if $\sum a_n$ and $\sum b_n$ are both absolutely convergent. – Taladris Aug 21 '23 at 10:17
  • A strategy that often works for series is to make a list of the tests you know and check whether the data of the problem satisfies the conditions of one (or more) of the tests. Here, there is no "standard test" (e.g. from Stewart's Calculus) whose conditions are satisfied, so the claim is probably false. – Taladris Aug 21 '23 at 10:22
  • 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 '23 at 09:22

1 Answers1

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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.

Ingix
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