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If $\sum _{n=1}^{\infty \:}a_{n\:}$ converges and $\sum _{n=1}^{\infty \:}b_{n\:}$ converges.
how to proof that $\sum _{n=1}^{\infty \:}a_{n\:}-b_n$ also converges?

user2637293
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    What do you know about the corresponding statement for sequences? Note an infinite sum converges if and only if its sequence of partial sums converges. – David Mitra Jun 05 '14 at 11:35
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    it's better to write $\sum_{n}(a_n -b_n)$ – Alex Jun 05 '14 at 12:05

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Lets call the infinite sum of $a$ the $S_a$, similiarly $S_b$ the sum of $b$

The sum of diferences can be expresed as the diference of two sums $$S_{a-b}=S_a-S_b$$

saying that a sum $S$ converges is the same as saying that $S<\infty$ so $S_a<\infty $ and $ S_b<\infty$

now without loss of generality, lets say that $S_a\geq S_b$

thus we can substract $S_a$ from $S_b$ in the inequality and it will still hold: $$S_b-S_a<\infty$$

cirpis
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  • that's if they are allowed to use the fact that you can interchange convergent sums. Usually in such cases students are expected to use the fact that $|s_n - L| \leq \epsilon $ for $ n > N$ – Alex Jun 05 '14 at 12:10