# Can we pull out a constant of a divergent series?

I know that if a series converges, the following applies:

$$\sum_{n=i}^\infty c a_n = c \sum_{n=i}^\infty a_n$$

However, I can't seem to find any info on whether this holds for diverging series as well. The property is often mentioned together with this one, of which I know it does not apply to divergent series:

$$\sum_{n=i}^\infty a_n + b_n= \sum_{n=i}^\infty a_n + \sum_{n=i}^\infty b_n$$

This makes me think the first property might require the same condition, but I'm not sure.

• Why would you want to? – Ali Caglayan Oct 31 '14 at 18:50
• Well, you can pull the constant out in the sense that the series on the left converges iff the constant times the series on the right converges, so it doesn't change things. That said, as Alizter said, it's not clear what the point is. – Reinstate Monica Oct 31 '14 at 18:52
• This only makes sense if you ask a question like: if $ca_n$ diverges does $a_n$ diverge? The answer is yes and can be most easily proven by assuming that it converges and deriving a contradiction. – Winther Oct 31 '14 at 18:53
• The reason the sum of two divergent series behaves unpredictably is that they might coincidentally cancel each other out (say, $a_n=n$, $b_n=-n$). But if you're multiplying each term of the sum by the same number, there's no way that can introduce cancelation, unless that number is zero... – Micah Oct 31 '14 at 18:54
• If $\sum a_n$ diverges, then $\sum_{n=i}^\infty a_n$ either isn't something that exists, or it's $\infty$. It's true that $\infty = \infty$, and otherwise the statement is meaningless. – Najib Idrissi Oct 31 '14 at 19:02

Let $\sum c a_n \to +\infty$.
That means $\forall M > 0 :\exists N > 0: n > N \implies \sum_{i=1}^n ca_n > M$
If $c > 0$, you can divide both sides of the last inequality by $c$ and you can put $M' = \frac M c$.