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suppose $\sum a_n$ converges. Is it true that then $\sum (-1)^na_n$ will also converge.

I think that the statement is true but I'm having trouble proving it.

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  • $\begingroup$ Do you know about absolute convergence? $\endgroup$ – Randall Apr 5 '18 at 19:44
  • $\begingroup$ so since it converges absolutely then $\sum (-1)^na_n$ will also converge? @Randall $\endgroup$ – Skrrrrrtttt Apr 5 '18 at 19:47
  • $\begingroup$ abs cov says if $\sum_n |a_n|$ converges then so does $\sum_n a_n$. That's exactly your situation. $\endgroup$ – Randall Apr 5 '18 at 19:48
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    $\begingroup$ @Randall that's not the situation because $a_{n}$ converging does not imply $|a_{n}|$ converging. $\endgroup$ – pwerth Apr 5 '18 at 19:50
  • $\begingroup$ @Randall $|a_n| \neq a_n$. $\endgroup$ – Mesmerized student Apr 5 '18 at 19:52
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Let consider

  • $a_n=(-1)^{n}\frac1n\implies \sum a_n$ converges

but

  • $\sum (-1)^na_n=\sum \frac1n$ which diverges
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No, this is in general wrong. Consider for instance $a_n = \frac{(-1)^n}{n}$.

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Harmonic series can be counterexample for your statement. So no, it’s not true.

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$a_k= \frac 1k$ if $k$ even $a_k= -\frac 1{k-1}$ is $k$ odd

$\sum (-1)^ka_k = \sum \frac 1{k}$ diverges but $\sum a_k = \frac1{2n} $ converges

Your result is true is all $a_k$ are positive in the first place or if the |a_k| serie converges

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