It is true to say that $\sum_{n=1}^{\infty} (a_n+b_n)=\sum_{n=1}^{\infty} a_n + \sum _{n=1}^{\infty} b_n$? It seems false but i can't find any counterexample..

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    $\begingroup$ $\sum (1+(-1)) =\cdots$ ?? $\endgroup$ Jun 5, 2014 at 12:39
  • 1
    $\begingroup$ Correct me if I'm wrong, but I believe it's correct if all series in question converge. $\endgroup$
    – Alex G.
    Jun 5, 2014 at 12:41
  • $\begingroup$ Any two of the series in the equation being convergent is enough... $\endgroup$
    – Macavity
    Jun 5, 2014 at 12:46
  • $\begingroup$ Pick any (bounded, for simplicity) sequence $a_n$ such that $\sum_{n=1}^\infty a_n$ diverges. Then pick $b_n = -a_n$. This generates plenty of counter examples. $\endgroup$ Jun 5, 2014 at 12:47
  • $\begingroup$ why its matter even thinking of convergence of these series? $\endgroup$ Jun 5, 2014 at 12:49

4 Answers 4


If $\sum a_n$ and $\sum b_n$ converge, then $\sum (a_n+b_n)$ converges to $\sum a_n+\sum b_n$.

If $\sum a_n$ converges but $\sum b_n$ diverges, or if $\sum a_n$ diverges and $\sum b_n$ converges, then $\sum (a_n+b_n)$ diverges.

If $\sum a_n$ and $\sum b_n$ diverge, then $\sum (a_n+b_n)$ may converge or diverge.

  • $\begingroup$ can you please prove your second statement? $\endgroup$ Jun 5, 2014 at 12:50
  • $\begingroup$ ∑(an+bn−an)=∑(an+bn)−∑bn why is this even true? $\endgroup$ Jun 5, 2014 at 12:57

One helpful point might be to think in terms of the partial sums. Let $A_N = \sum\limits_{n=1}^N a_n$, $B_N = \sum\limits_{n=1}^N b_n$, and $C_N = \sum\limits_{n=1}^N (a_n + b_n)$. It is clear for every $N \in \Bbb{N}$ that $A_N + B_N = C_N$. Now, let's apply the theory of sequences.

You know that if $A_N$ converges as $N \to \infty$ and $B_N$ converges as $N \to \infty$ then $A_N + B_N$ converges and $\lim (A_N + B_N) = \lim A_N + \lim B_N$, therefore $\lim C_N = \lim A_N + \lim B_N$.

The other cases can be handled similarly. Such as if $A_N \to \infty$ and $B_N$ converges then $A_N + B_N \to \infty$.


A counter example is the following:

$$ a_n=\dfrac{1}{n}, $$ $$ b_n=\dfrac{-1}{n+1}, $$

So you have:

$$ \begin{equation} \begin{split} \sum_{n=1}^{\infty}\left(a_n+b_n\right)&=\sum_{n=1}^{\infty}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right),\\ &=\underbrace{\sum_{n=1}^{\infty}\left(\dfrac{1}{n(n+1)}\right)}_{\mathrm{converges}},\\ &\neq \sum_{n=1}^{\infty}a_n+\sum_{n=1}^{\infty}b_n,\\ &\neq \underbrace{\sum_{n=1}^{\infty}\dfrac{1}{n}}_{\mathrm{diverges}:\,\\\mathrm{Harmonic\; series}}-\underbrace{\sum_{n=1}^{\infty}\dfrac{1}{n+1}}_{\mathrm{diverges}:\,\\\mathrm{Harmonic\; series}}. \end{split} \end{equation} $$

  • $\begingroup$ yes but i did'nt actually asked about the convergence of the series...just if the equation is true $\endgroup$ Jun 5, 2014 at 14:18
  • $\begingroup$ It is a counter example that shows that it is not true. Because if it was true, we would not have $\infty$-$\infty$ after the split of the sum. $\endgroup$
    – Jika
    Jun 5, 2014 at 14:22
  • $\begingroup$ no it's not..its counter example to the sentence: if an+bn converge then an converge and bn converge. NOT to the question i asked.. $\endgroup$ Jun 5, 2014 at 14:25
  • $\begingroup$ See the first comment of David Mitra. $\endgroup$
    – Jika
    Jun 5, 2014 at 14:32
  • $\begingroup$ @user2637293 You asked whether the LHS and RHS are equal. If the LHS is defined and the RHS is not, then they are not equal. Thus this is a completely legitimate answer to your question. $\endgroup$
    – user21467
    Jun 5, 2014 at 21:05

Actually is true if $\ \sum _{n=1}^{\infty }\:a_n$ and $\ \sum _{n=1}^{\infty }\:b_n$ converges.

For example, let $a_{n}=1$ and $b_{n}=-1$ for all $n\in \mathbb{N}$, then $\ \sum _{n=1}^{\infty }\:a_n=\ \sum _{n=1}^{\infty }\:1$ and $\ \sum _{n=1}^{\infty }\:b_n=\ \sum _{n=1}^{\infty }\:-1$ doesn't converge, but $\ \sum _{n=1}^{\infty }\:a_n+b_n=\ \sum _{n=1}^{\infty }\:0=0$ converges.

The answer provided by Jasper should give you the reason. Just write if you need a proof.


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