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I'm trying to prove that if $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ are convergent series and $c_n = \max\{a_n, b_n\}$ then $\sum_{n=1}^\infty c_n$ doesn't necessarily converge.

I believe I've come up with a counter-example but to be honest I'm not too confident in my ability to work with series so I was wondering if I could get some feedback.

Let $\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{(-1)^n}{n}$ and $\sum_{n=1}^\infty b_n = \sum_{n=1}^\infty \frac{1}{n^2}$.

These series both converge but $c_n = \frac{1}{n}$ if $n$ is even and $c_n = \frac{1}{n^2}$ if $n$ is odd, so I don't think $\sum_{n=1}^\infty c_n$ converges.

The reason I think this is because $\sum_{n=1}^\infty c_{2n} = \sum_{n=1}^\infty \frac{1}{2n} = \frac{1}{2}\sum_{n=1}^\infty \frac{1}{n}$ which diverges since it's a multiple of the harmonic series. Since the $c_n$ terms with odd indices are all positive, the sum of the two should be divergent as well.

Could anyone confirm I'm thinking about this correctly or point out a flaw in my reasoning if I'm not?

Thank you

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    $\begingroup$ This looks good. $\endgroup$ Mar 4 '15 at 16:46
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Your proof looks good. Even simpler counterexample is $a_n=\frac{(-1)^n}{n}$ and $b_n=-a_n$.

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