# What are the rules for convergence for 2 series that are added/subtracted/multiplied/divided?

Are there general rules that apply? For example:

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

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

etc.

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

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

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).

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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 $y_n=n^2$.

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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