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
if any one can explain or give me a link for learning about this kind of sumation problems.
Advanced thanks for your help
 A: 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.

A: 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 $M\gt{N}$:
$$\sum_{k=0}^Ma_k+\sum_{k=0}^N{b_k}=\sum_{k=0}^{N}(a_k+b_k)+\sum_{k=N+1}^Ma_k$$
A: 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
$$
- \frac 1 2 - \frac 1 4 - \frac 16 - \cdots = -\infty.
$$
A: 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.
