A constant and divergent series

Question

Let $\sum_{n = 1}^\infty 1/3n$ , can we rewrite it as $1/3 \times \sum_{n = 1}^\infty 1/n$ and then conclude it's divergent ? It's true for convergent sequences but what about divergent sequences ? Intuitively , it seems reasonable to me but I don't know how to prove it .

Answer 1

Let $u_{n}=\displaystyle\sum_{k=1}^{n}\dfrac{1}{3k}$, then $3u_{n}=\displaystyle\sum_{k=1}^{n}\dfrac{1}{k}\rightarrow\infty$, in other words, $3u_{n}\rightarrow\infty$. Now we prove that $u_{n}\rightarrow\infty$:

Given $M>0$, there exists some positive integer $N$ such that for all $n\geq N$, $3u_{n}>3M$, then $u_{n}>M$ for all such $n$, this proves that $u_{n}\rightarrow\infty$.