# whether $\sum a+\sum b=\sum (a+b)$ is correct or not

i am a chemist and I am doing some calculations when i came across this problem.

$\sum a+\sum b=\sum (a+b)$ ;is this equation true

and

$\sum a_{n}+\sum b_{n}=?$ ;what is the answer for sum of n number of a and b are added

and

$\sum a_{m}+\sum b_{n}=?$ ;what is the answer for sum of m number of a and n number of b were added

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The last question is:$$\sum_0^m{a_k}+\sum_0^n{}b_k$$. Is it correct? – Riccardo.Alestra Feb 23 '12 at 11:55
You may try to change indices if you get stuck with different indices m and n. Then, you can easily sum up these sequences which have the same indices. Of course, you should arrange the beginning and end points, i.e. see the answer below. – Ragnar Feb 23 '12 at 12:32
Is addition of finitely many terms commutative? Yes. Some caution is required when summing infinite series, however... – Zhen Lin Feb 23 '12 at 12:44

There is law that says if you have infinite conditionally convergent sum of real numbers then you can shuffle its components so it will converge to any number you want. So if you would like to ask little more generic question answer will be that you can not shuffle sum.

Here is example Riemann series theorem

There is also one more interesting fact according to wolfram math when series is absolute convergent the result of summation is independend on order of summation. There is also article on wikipedia on unconditional convergence -- this is kind of convergence that allows to rearange terms.

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I think it may be a good idea to emphasise what the conditional convergence is, since the OP is not a mathematician. – savick01 Feb 24 '12 at 11:28
In artickle that is linked to my answer there is link to description of what conditionally convergence is but I think wikipedia has much more informative article. – Trismegistos Feb 24 '12 at 17:23
Great, a paragraph about absolute convergence is just what I felt was missing here but I didn't realise that! – savick01 Feb 24 '12 at 17:56

It is true when the two sums have only finitely many terms.

With infinitely many terms, it is true if the sums are absolutely convergent, i.e. the sums of the absolute values converge.

In other words, if $$\sum_n |a_n| <\infty\text{ and }\sum_n|b_n| < \infty.$$

If the sums of absolute values are finite, then the sums without the absolute values in some cases converge, but the number they converge to can change if the order in which they are added is changed, and then your two sums can be different.

Probably the most well known example is the alternating harmonic series: $$1 - \frac 1 2 + \frac 1 3 - \frac 1 4 + \frac 1 5 - \frac 1 6 + \cdots = \log_e 2,$$ but $$1 + \frac 1 3 + \frac 1 5 + \frac 1 7 + \cdots = \infty$$ and $$- \frac 1 2 - \frac 1 4 - \frac 16 - \cdots = -\infty.$$

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What about $\sum_n \left((-1)^n + (-1)^{n-1}\right)$? – JavaMan Feb 23 '12 at 16:21
It's also the case that if $\sum a_n$ and $\sum b_n$ exist and are finite, then their sum is equal to $\sum (a_n+b_n)$, regardless of whether the first two converge absolutely or conditionally. – anon Feb 24 '12 at 18:00

If $N=M$ you can write $$\sum_{k=0}^Ma_k+\sum_{k=0}^N{b_k}=\sum_{k=0}^{N}(a_k+b_k)$$ If $M\gt{N}$: $$\sum_{k=0}^Ma_k+\sum_{k=0}^N{b_k}=\sum_{k=0}^{N}(a_k+b_k)+\sum_{k=N+1}^Ma_k$$

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In the last line, your last sum should start at $N+1$. – M Turgeon Feb 23 '12 at 16:06

I would like to stress one additional thing which is not included in the above answers.

Assuming that $\sum_{i=1}^\infty a_i$ and $\sum_{i=1}^\infty b_i$ are convergent to say $A$ and $B$, we know that $\sum_{i=1}^\infty (a_i + b_i)$ is convergent and its limit is $A+B$. So if we only know that the problem of finding $\sum_{i=1}^\infty a_i + \sum_{i=1}^\infty b_i$ is well defined (series are convergent), then we have: $$\sum_{i=1}^\infty a_i + \sum_{i=1}^\infty b_i = \sum_{i=1}^{\infty} (a_i+b_i).$$

A little bit more is also true: if $\sum a_i$ and $\sum b_i$ diverges to infinity or converges to some value AND the sum $A+B$ can be defined (ie. it is not $\infty - \infty$ or sth similar. $-\infty - \infty=-\infty$, $-5 + \infty = \infty$ etc. are OK), then the above is also true.

EDIT

Since I was downvoted, I think I need to give a proof.

First for convergent series. $A$ is said to be the value of $\sum_{i=1}^\infty a_i$ if $\lim_{n \to \infty} \sum_{i=1}^n a_i = A$. Let $\varepsilon>0$. From the assumption we have $N$ satisfying: for $n>N$ $|\sum_{i=1}^n a_i - A| < \frac{\varepsilon}{2}$ and the same for $\sum b_i$. So we have that for $n>N$: $$|\sum_{i=1}^n (a_i+b_i) - (A+B)| = |\sum_{i=1}^n a_i + \sum_{i=1}^n b_i - (A+B)| =$$ $$= |\sum_{i=1}^n a_i - A + \sum_{i=1}^n b_i - B| \leq |\sum_{i=1}^n a_i - A| + |\sum_{i=1}^n b_i - B| \lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$

Second - if one of them (say $\sum a_i$) diverges to infinity and the other diverges to infinity or converges to a finite value. (The case with minus infinity/infinities is the same.) Take any $M < \infty$. If $\sum_{i=1}^\infty b_i$ is $\infty$, then $k=0$. Otherwise $k=\sum_{i=1}^\infty b_i$. Let $P=M-k+1$.

From the assumptions we know that there exists $N$ such that for $n>N$ we have: $\sum_{i=1}^n a_i > P$ and $\sum_{i=1}^n b_i > k-1$. Consequently $\sum_{i=1}^n (a_i+b_i) = \sum_{i=1}^n a_i + \sum_{i=1}^n b_i > P + k-1 = M$, qed.

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@MichaelHardy: Are there any gaps in the above proofs? – savick01 Feb 23 '12 at 19:15
@MichaelHardy: I don't understand your comment, can you elaborate? I think Savicko1's original answer (before any revisions) was perfectly correct. If $\sum_n a_n =A$ and $\sum_{n} b_n =B$ then $\sum_{n} (a_n+b_n)=A+B$; if you have any two converging you automatically get the third. Perhaps you should be a bit more careful before making false accusations about new users answers... – Eric Naslund Feb 23 '12 at 19:57