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Let $a,b \in \mathbb{R}$, $(x_{n})_{n}$ and $(y_{n})_{n}$ such that $x_{n} > 0$ and $y_{n} > 0$ for all $n$ in $\mathbb{N}$.

If $\sum_{n=0}^{\infty} ax_{n} + by_{n} = c \in \mathbb{R}$, can we say that

$\sum_{n=0}^{\infty} ax_{n} + by_{n} = \sum_{n=0}^{\infty} ax_{n} + \sum_{n=0}^{\infty} by_{n} = c$ ?

Thanks in advance

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    Only if both of the series are convergent by themselves. – 2017-01-09
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    @LaarsHelenius I guess that if $\sum (ax_{n} + by_{n}) < +\infty$ then both series converges , because of they're positive. – 2017-01-09
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    @openspace Not necessarily, because $a,b\in\Bbb R$. – 2017-01-09
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    @G.Sassatelli yes, my bad. Forgot about $a$ and $b$. – 2017-01-09

2 Answers 2

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No, in general we cannot say that. The problem is, that from the convergence of $$ \sum_{n=0}^\infty ax_n + by_n $$ we cannot infer the convergence of $\sum_{n=0}^\infty x_n$. To give an example, let $$ x_n = \frac 1{n+1}, \qquad y_n = \frac 1{n+1} $$ $a = 1$ and $b = -1$. Then $ax_n + by_n = 0$, hence $$ \sum_{n=0}^\infty ax_n + by_n = \sum_{n=0}^\infty 0 = 0. $$ But $\sum_{n=0}^\infty x_n = \sum_{n=0}^\infty \frac 1n$ does not exist.

If on the other hand $\sum_{n=0}^\infty x_n$ and $\sum_{n=0}^\infty y_n$ both exist, then $$ \sum_{n=0}^\infty ax_n + by_n = a \sum_{n=0}^\infty x_n + b\sum_{n=0}^\infty y_n. $$

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Take $a=1, b=-1$

$$x_n=\frac{1}{(n+1)^2}+\frac{1}{n+1}$$

$$y_n=\frac{1}{n+1}.$$

and conclude.