Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Are there general rules that apply? For example:

(convergent series) + (divergent series) = (divergent series)

(convergent series) * (divergent series) = (convergent series)


Are there steadfast rules like this? Or does it vary depending on the specific series?

share|improve this question
For multiplication, you may be interested in Dirichlet's test for convergence. –  Antonio Vargas May 19 '12 at 7:45

2 Answers 2

For the addition : let $\sum u_n$, $\sum v_n$ and $\sum (u_n+v_n)$.

  • If two of these series converge the last converges.
  • If one converges and another diverges then the last diverges.
  • If two diverge, we can't say anything about the last. (e.g. $\sum n$ and $\sum (-n)$).

You can easily adapt this for the substraction.

For the Cauchy product :

  • If $\sum u_n$ and $\sum v_n$ are absolutely convergent then the Cauchy product is absolutely convergent.
  • If one is absolutely convergent and the other convergent, the Cauchy product is convergent (Mertens' theorem).
  • If they are (conditionally) convergent you can't say anything (e.g. $u_n=v_n=\frac{(-1)^n}{\sqrt{n+1}}$ and $u_n=v_n=\frac{(-1)^n}{n+1}$).
  • Nevertheless, if $\sum u_n$, $\sum v_n$ and their Cauchy product are convergent, then the Cauchy product is equal to $(\sum u_n)(\sum v_n)$.
  • We can't say anything about the Cauchy product of divergent series (Think of power series).

A quick summary :
If $E$ is a Banach space (to have absolute convergence $\Rightarrow$ convergence) : the set $\mathcal S_C$ of convergent series with terms in $E$ is a vector space and the set $\mathcal S_{AC}$ of absolutely convergent series is a vector subspace of $S_C$.
Moreover, if $E=\mathbb R$ or $E=\mathbb C$, $S_{AC}$ has a ring structure.

If you want to consider $\sum u_nv_n$ instead of the Cauchy product you can use the Dirichlet's test or an Abel transformation.

A strange operation is to permut terms : if $\sum u_n$ is absolutely convergent then $\sum u_{\varphi(n)}$ too, and $\sum u_n=\sum u_{\varphi(n)}$. But if $\sum u_n$ is conditionally convergent the result is false. Worse, $\forall S\in\mathbb R\cup\{+\infty,-\infty\}$, you can find a permutation $\varphi$ such as $\sum u_{\varphi(n)}=S$.

Another useful operation is to group terms : let $(p_n)_{n\in\mathbb N}$ a (strictly) increasing sequence with $p_n\in\mathbb N$. Let $v_0=\displaystyle\sum_{i=0}^{p_0}u_i$ and $v_n=\displaystyle\sum_{i=p_{n-1}+1}^{p_n}u_i$.
If $\sum u_n$ converges, then $\sum v_n$ converges too and $\sum u_n=\sum v_n$.
But, we usually can't say anything if $\sum u_n$ diverges : take $u_n=(-1)^n$ and define $v_n$ by grouping two following terms $v_n=1-1=0$, then $u_n$ diverges but $v_n$ converges.
(There are still some results, but the answer would be too long).

Conclusion : take care when manipulating series. If I remember well, Euler made this error : let $S=1-1+1-1+\cdots=1-(1-1+1-\cdots)=1-S\Rightarrow S=\frac{1}{2}$ (that would be true if we used the Cesaro limit of partial sums) or maybe $S=1+2+4+8+\cdots=1+2(1+2+4+\cdots)=1+2S\Rightarrow S=-1$ (it would be true for the field of 2-adic numbers, but Newton didn't know that).

share|improve this answer

The first rule you mention is true, as if $\sum x_n$ converges and $\sum y_n$ diverges, then we have some $N\in\mathbb N$ such that $\left|\sum\limits_{n=N}^\infty x_n\right|<1$ so $\left|\sum\limits_{n=N}^\infty (x_n+y_n)\right|\geq \left|\sum\limits_{n=N}^\infty y_n\right|-1\to\infty$. However, the second is false, even if the series converges to $0$. An easy example is when $x_n=1/n^2$ and $y_n=n^2$.

share|improve this answer
so is there a set of properties like this somewhere? i couldn't find them. I want to be able to look at a series and say, "ok i can split this up into two series, see if each converges, and use that to decide whether the original series converges." –  Marty Apr 18 '12 at 9:37
@FrederickCraine It sounds like the concept of absolute convergence is really what you're after; you'll notice from JBC's answer that it obviates most of the issues of splitting, rearrangement of terms, etc. that you might have... –  Steven Stadnicki Jun 19 '12 at 21:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.