# Sum of two divergence series is always divergence series?

In my opinion, I still believe that sum of two divergence series can be convergence series.

But here is my proof:

First, we have an obvious result: sum of two convergence series is convergence series.

Take opposite of this statement is: if exist one series is (non-convergence) so cannot be (non-convergence)

$==>$ if exist one series is divergence, so sum always be divergence.!!! (because non-convergence series is divergence series and vice verse)

If my proof is wrong, please tell me where and give me an example that sum of two divergence series can be convergence series please.

Thanks :)

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Take $\sum{1\over n}$ and $\sum{-1\over n}$ (there are less trivial examples). It's true if both series consist of nonnegative terms. – David Mitra Jun 3 '12 at 16:58
@DavidMitra yes. I have thought your example before, but it's too special. – hqt Jun 3 '12 at 17:00
What do you actually mean by adding two series? Do you mean $$\left(\sum_{n=1}^{\infty} a_n \right) + \left(\sum_{n=1}^{\infty} b_n \right)$$ or $$\sum_{n=1}^{\infty} \left(a_n + b_n \right) ?$$ – user17762 Jun 3 '12 at 17:00
Too special for what exactly? – anon Jun 3 '12 at 17:00
Take opposite of series and plus together is too special in my opinion. – hqt Jun 3 '12 at 17:03

If $a_n+b_n = c_n$ and $\sum_n c_n$ is a convergent series, then $b_n=c_n-a_n$, so $\sum_n b_n$ differs from $\sum_n (-a_n)$ by the convergent series $\sum_n c_n$. In particular, $\sum_n b_n$ is convergent if and only if $\sum_n a_n$ is.

If you start with a particular divergent series $\sum_n a_n$, every series you can add termwise to $\sum_n a_n$ to obtain a convergent series has the form $\sum_n(-a_n+c_n)$, where $\sum_n c_n$ is whatever convergent series you like.

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The proof is wrong and here is why. The theorem states that if we add two convergent series then we get a new convergent series. So that statement is equivalent to the following. If a divergent series was produced by sum of two series then these two series were not convergent series (so at least one is divergent).This is different than saying adding two divergent series produces a divergent series. It is like having A implies B this is indeed equivalent to notB implies notA but is certainly different from notA implies notB.

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I agree that the sum of two divergent series can be convergent.
Consider the infinite series of $a(n)=$ sum of $1=1+1+1+1\rightarrow \infty$
$b(n)=$ sum of $-1=-1+-1+-1 \rightarrow -\infty$
The first term of the series $a(1)+b(1)=1-1=0$
The second term of the series $a(2)+b(2)=1-1=0$
and so forth ....
Therefore all the partial sums are $0$ and the limit of the $n^{th}$ partial sum is $0$ and so my the divergent series converge to $0$

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Given the simplicity of this example, I suspect you did not thoroughly read the earlier comments and answers, or even the Question itself. If you had, you surely would have remarked on the incorrectness of the "proof" offered in the (three year old) Question (although your example is correct, the OP comments on the Question itself that such an example is "too special", whatever that means). – hardmath Apr 9 at 17:27