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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?

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

1 Answer 1

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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.

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