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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asked Jun 5, 2014 at 12:39
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5$\begingroup$ $\sum (1+(-1)) =\cdots$ ?? $\endgroup$– David MitraJun 5, 2014 at 12:39
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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
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$\begingroup$ Any two of the series in the equation being convergent is enough... $\endgroup$– MacavityJun 5, 2014 at 12:46
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$\begingroup$ 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. $\endgroup$– Nicholas StullJun 5, 2014 at 12:47
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$\begingroup$ why its matter even thinking of convergence of these series? $\endgroup$– user2637293Jun 5, 2014 at 12:49
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4 Answers
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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.
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$\begingroup$ can you please prove your second statement? $\endgroup$ Jun 5, 2014 at 12:50
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$\begingroup$ ∑(an+bn−an)=∑(an+bn)−∑bn why is this even true? $\endgroup$ Jun 5, 2014 at 12:57
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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 $A_N + B_N \to \infty$.
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A counter example is the following:
$$ a_n=\dfrac{1}{n}, $$ $$ b_n=\dfrac{-1}{n+1}, $$
So you have:
$$ \begin{equation} \begin{split} \sum_{n=1}^{\infty}\left(a_n+b_n\right)&=\sum_{n=1}^{\infty}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right),\\ &=\underbrace{\sum_{n=1}^{\infty}\left(\dfrac{1}{n(n+1)}\right)}_{\mathrm{converges}},\\ &\neq \sum_{n=1}^{\infty}a_n+\sum_{n=1}^{\infty}b_n,\\ &\neq \underbrace{\sum_{n=1}^{\infty}\dfrac{1}{n}}_{\mathrm{diverges}:\,\\\mathrm{Harmonic\; series}}-\underbrace{\sum_{n=1}^{\infty}\dfrac{1}{n+1}}_{\mathrm{diverges}:\,\\\mathrm{Harmonic\; series}}. \end{split} \end{equation} $$
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$\begingroup$ 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
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$\begingroup$ 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. $\endgroup$– JikaJun 5, 2014 at 14:22
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$\begingroup$ 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
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$\begingroup$ @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. $\endgroup$– user21467Jun 5, 2014 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.
