Prove or disprove:

If $$\sum_{k=1}^\infty a_{2k}, \sum_{k=1}^\infty a_{2k-1}$$ are convergent series, then $$\sum_{n=1}^\infty a_n $$ also converges.

$a_{2k} $ and $a_{2k-1}$ are sub-sequences of $a_n$ I know that the statment is correct, but I don't know how to prove it.

I am especialy "afraid" changing the series elements as I know it's not allowed.

Any ideas?


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    $\begingroup$ Call the even terms $b_k$ and the odd terms $c_k$. Consider the sums $b_1+0+b_2+0+\cdots$ and $0+c_1+0+c_2+\cdots$. $\endgroup$ – David Mitra Jun 2 '17 at 11:54

Let $(S_n)$ be the sequence of partial sums of $\sum a_n$. We have:

$$S_{2n} = \sum_{k=1}^{n} a_{2k} + \sum_{k=1}^{n}a_{2k-1}$$


$$S_{2n-1} = \sum_{k=1}^n a_{2k-1} + \sum_{k=1}^{n-1} a_{2k}$$

Thus, $(S_{2n})$ and $(S_{2n-1})$ converge to the same limit. Hence, $(S_n)$ is convergent and so is $\sum a_n$.

  • $\begingroup$ Intuitively I can understand why it's correct, but how can I prove that it's allowed to do that? I have changed the order of the elements in my series, didn't I? $\endgroup$ – Alan Jun 2 '17 at 11:55
  • $\begingroup$ @Alan allowed to do what? Is there something non-rigorous in my answer? $\endgroup$ – user384138 Jun 2 '17 at 11:55
  • $\begingroup$ First of all, thank you. But can you please explain why $S_{2n-1}$ equals what you wrote, and $S_{2n}$ as well? $\endgroup$ – Alan Jun 2 '17 at 11:59
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    $\begingroup$ @Alan you're welcome! It's intuitively clear, no? As for rigour, you can prove it by induction, for instance. $\endgroup$ – user384138 Jun 2 '17 at 11:59
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    $\begingroup$ @Alan it's $n$, just as written. $\endgroup$ – user384138 Jun 2 '17 at 12:32

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