# Sum of two conditionally convergent series

Suppose I have two conditionally convergent series $\sum \limits_{n=1}^{\infty} s_n$ and $\sum \limits_{n=1}^{\infty} t_n$.
According to http://mathworld.wolfram.com/ConvergentSeries.html the series $\sum \limits_{n=1}^{\infty} s_n + t_n$ will then also be convergent. Does this hold for conditionally convergent series? (I just want to be sure :D) Also, if it holds, can one say anything about the value of $\sum \limits_{n=1}^{\infty} s_n + t_n$?
Thanks

• Well, $\sum_{n=1}^\infty s_n + t_n = \sum_{n=1}^\infty s_n + \sum_{n=1}^\infty t_n$, doesn't it? – Andy Nov 1 '11 at 11:02
• If the series were absolutely convergent, yes. But since a reordering of terms in conditionally convergent series may change the limit this is (imho) not trivial. – iolo Nov 1 '11 at 11:21
• see them, as the answer says, as limits of a sequence, it gets easier :) – Andy Nov 1 '11 at 11:29
• I don't understand. You have two conditionally convergent series. They satisfy a certain property. Then you ask if this certain property holds for conditionally convergent series. – zhw. Apr 20 '16 at 2:37

This is a special case of the general property of sequences that the sum of two convergent series converges to the sum of their limits.

• Hm... I guess i have to take that, although I'd love to see a proof for that property ;) – iolo Nov 1 '11 at 11:25
• @iolo. Suppose $a_n \to a$ and $b_n \to b.$ Then by definition, for any $\epsilon > 0$ we can find $N \in \mathbb{N}$ such that $|a- a_n| < \epsilon/2$ for all $n > N.$ Similarly, we can find $M\in\mathbb{N}$ such that $|b-b_n| < \epsilon/2$ for all $n>M.$ Thus, for all $n> \max \{N,M \}$ we have $|(a+b) - (a_n + b_n) |= |(a-a_n) + (b-b_n) | \leq |a-a_n| + |b-b_n| < \epsilon$, so $a_n + b_n \to a+b.$ – Ragib Zaman Nov 1 '11 at 11:43
• Nice. Clean and short. Thank you :) – iolo Nov 1 '11 at 11:58

If I may speculate, I think you are the latest student to get confused by the term conditionally convergent. Recall that by definition a conditionally convergent series is one which is convergent but not absolutely convergent...so in particular every conditionally convergent series is convergent! Thus everything which holds for all convergent series holds for conditionally convergent series, and this definitely applies to your question.

The relevant fact here is that the sum of two convergent series is a convergent series. This is not a proof you need to supply on your own: it should be found in absolutely (!) every introductory text in which infinite series are discussed. For instance, it is Proposition 39a) in these notes.

In the interests of both edification and entertainment, let me attempt to add a rider to your question: if $\sum_n a_n$ and $\sum_n b_n$ are both conditionally convergent series, we know that $\sum_n a_n + b_n$ must converge. But can it converge absolutely?

1) If no sign restrictions are placed on $a_n$ and $b_n$, then the answer is yes. For instance, we can take any conditionally convergent series $\sum_n a_n$ (the alternating harmonic series is a popular one) and put $b_n = -a_n$ for all $n$.

2) This makes one wonder what happens if for all $n$, the terms $a_n$ and $b_n$ have the same sign. Here the answer is no: the series cannot converge absolutely if even one of $\sum_n a_n$, $\sum_n b_n$ converges nonabsolutely. This follows quickly from the fact that in a series which converges nonabsolutely, both the series of positive parts and the series of negative parts must diverge ($\S 8.4$ of the same notes).