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If $\sum a_n$ is convergent and $a_n>0$, then $\sum (-1)^n a_n$ is convergent.

So far I've tried convergence/divergence tests and also I tried to prove this using partial sums. But tests do not work because the latter series includes negative terms. Maybe the Alternating Series Test could have worked but "$a_n$ is decreasing" is not an initial condition. I know the proof will be very easy but I am stuck here. Could you provide me with a little hint?

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    $\begingroup$ The hint of Nicolò Ruggeri is the good one. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 23 '16 at 14:02
  • $\begingroup$ The sum of odd and even terms (separately) converge, so you can combine them together into another convergent sequence. $\endgroup$ – A.S. Feb 23 '16 at 14:03
  • $\begingroup$ $\sum_{n \le N} a_{2n} \le \sum_{n \le 2N} a_{n}$ so $\sum_n a_{2n}$ converges $\endgroup$ – reuns Feb 25 '16 at 7:53
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HINT: any absolutely convergent series is convergent.Let me know if you want the full solution

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    $\begingroup$ That is another theorem to prove. $\endgroup$ – frosh Feb 23 '16 at 14:05
  • $\begingroup$ This would be the main way to do this, but if you don't know the theorem try with A.S hint, it works too! $\endgroup$ – Nicolò Feb 23 '16 at 14:08
  • $\begingroup$ Both has come to my mind but I was under the impression that there is one very easy answer. $\endgroup$ – frosh Feb 23 '16 at 14:09
  • $\begingroup$ Yes, as many times it's easy if you know the right theorem. Try looking on your book for mine, you'll study it for sure anyway! $\endgroup$ – Nicolò Feb 23 '16 at 14:11
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Theorem: If $\sum |a_n|$ is convergent, then $\sum a_n$ is convergent.

Proof: Let $b_n=|a_n|+a_n \Rightarrow b_n=0 \: \: \mbox{or} \: \: b_n=2|a_n| \Rightarrow 0\le b_n \le 2|a_n| \stackrel{\mbox{DCT}}{\Rightarrow} \sum b_n$ is convergent. Therefore, $\sum b_n - |a_n|=\sum a_n$ is convergent.

This answers the question I asked.

Remark: Actually, the result is known as Absolute Convergence Test. I am learning the subject this semester. My apologies for asking the question early.

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