# Is this proof correct? proving a sum of a convergent and divergent sequences is a divergent sequence

I'm trying to prove that if ${a_n}$ is convergent, and ${b_n}$ is divergent, then ${a_n} + {b_n}$ is a divergent series.

A proof a friend told, but I don't understand how can be correct is:

Assume ${a_n} + {b_n}$ convereges.

Therefore, $\lim_{n\to{\infty}}({a_n} + {b_n}) = \lim_{n\to{\infty}}({a_n}) + \lim_{n\to{\infty}}({b_n})$

Contradiction, since ${b_n}$ doesn't converge to a limit.

Why can limit arithmetic rules be used in this case? since ${b_n}$ is not a convergent series, I didn't think you can use it to express a convergent sequence's limit, and show "as if" ${b_n}$ has a limit and contradict the assumption using it.

Thanks.

• use $b_n=(a_n+b_n)-a_n$ – R.N Nov 17 '15 at 22:10

Rearrange the calculation so that you’re working with convergent sequences. Assume that $\lim\limits_{n\to\infty}(a_n+b_n)=L$ and $\lim\limits_{n\to\infty}a_n=M$; then
$$\lim_{n\to\infty}b_n=\lim_{n\to\infty}\big((a_n+b_n)-a_n\big)=\lim_{n\to\infty}(a_n+b_n)-\lim_{n\to\infty}a_n=L-M\;,$$
contradicting the assumption that $\langle b_n:n\in\Bbb N\rangle$ is divergent.
• @Taru: It’s okay because if the $a_n$ and $a_n+b_n$ sequences are convergent, which is what we’re assuming, then the displayed calculation is necessarily correct. Since the result of that calculation contradicts the assumptions, it follows that the $a_n$ and $a_n+b_n$ sequences cannot both be convergent. We know that one of them is, so we are forced to conclude that the other isn’t. – Brian M. Scott Nov 17 '15 at 22:39