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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..

user2637293
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    $\sum (1+(-1)) =\cdots$ ?? – David Mitra Jun 05 '14 at 12:39
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    Correct me if I'm wrong, but I believe it's correct if all series in question converge. – Alex G. Jun 05 '14 at 12:41
  • Any two of the series in the equation being convergent is enough... – Macavity Jun 05 '14 at 12:46
  • 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. – Nicholas Stull Jun 05 '14 at 12:47
  • why its matter even thinking of convergence of these series? – user2637293 Jun 05 '14 at 12:49
  • As in my hint to your previous question, you should think about series as limits of sequences, it's what they are. And you should know the answer regarding limits of sequences. – Git Gud Jun 05 '14 at 12:49
  • I think the exact answer to your question is 'it's not true in general' – Alex Jun 05 '14 at 15:01
  • @user2637293: You should realise that stating an identity implicitly implies one is saying that one side is defined if and only if the other side is defined. It is that aspect which fails here, and this is why thinking about convergence is necessary (and matters). – Marc van Leeuwen Jun 05 '14 at 15:14

4 Answers4

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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.

please delete me
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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 $A_N + B_N \to \infty$.

Tom
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A counter example is the following:

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

So you have:

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

Jika
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  • yes but i did'nt actually asked about the convergence of the series...just if the equation is true – user2637293 Jun 05 '14 at 14:18
  • 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. – Jika Jun 05 '14 at 14:22
  • 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.. – user2637293 Jun 05 '14 at 14:25
  • See the first comment of David Mitra. – Jika Jun 05 '14 at 14:32
  • @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. –  Jun 05 '14 at 21:05
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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.

Daniel
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